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Bram Gotink edited this page Oct 31, 2018 · 97 revisions

FAQs

Common "Bugs" That Aren't Bugs

I've found a long-overlooked bug in TypeScript!

Here are some behaviors that look may look like bugs, but aren't.

  • These two empty classes can be used in place of each other
  • I can use a non-void-returning function where one returning void is expected
  • I'm allowed to use a shorter parameter list where a longer one is expected
    • See the FAQ Entry on this page
    • Prior discussion at #370, #9300, #9765, #9825, #13043, #16871, #13529, #13977, #17868, #20274, #20541, #21868
  • private class members are actually visible at runtime
    • See the FAQ Entry on this page for a commonly suggested "fix"
    • Prior discussion at #564, #1537, #2967, #3151, #6748, #8847, #9733, #11033

Common Feature Requests

I want to request one of the following features...

Here's a list of common feature requests and their corresponding issue. Please leave comments in these rather than logging new issues.

  • Safe navigation operator, AKA CoffeeScript's null conditional/propagating/propagation operator, AKA C#'s' ?. operator #16
  • Minification #8
  • Extension methods #9
  • Partial classes #563
  • Something to do with this #513
  • Strong typing of Function members call/bind/apply #212
  • Runtime function overloading #3442

Type System Behavior

What is structural typing?

TypeScript uses structural typing. This system is different than the type system employed by some other popular languages you may have used (e.g. Java, C#, etc.)

The idea behind structural typing is that two types are compatible if their members are compatible. For example, in C# or Java, two classes named MyPoint and YourPoint, both with public int properties x and y, are not interchangeable, even though they are identical. But in a structural type system, the fact that these types have different names is immaterial. Because they have the same members with the same types, they are identical.

This applies to subtype relationships as well. In C++, for example, you could only use a Dog in place of an Animal if Animal was explicitly in Dog's class heritage. In TypeScript, this is not the case - a Dog with at least as many members (with appropriate types) as Animal is a subtype of Animal regardless of explicit heritage.

This can have some surprising consequences for programmers accustomed to working in a nominally-typed language. Many questions in this FAQ trace their roots to structural typing and its implications. Once you grasp the basics of it, however, it's very easy to reason about.

What is type erasure?

TypeScript removes type annotations, interfaces, type aliases, and other type system constructs during compilation.

Input:

var x: SomeInterface;

Output:

var x;

This means that at run-time, there is no information present that says that some variable x was declared as being of type SomeInterface.

The lack of run-time type information can be surprising for programmers who are used to extensively using reflection or other metadata systems. Many questions in this FAQ boil down to "because types are erased".

Why are getters without setters not considered read-only?

I wrote some code like this and expected an error:

class Foo {
   get bar() {
     return 42;
   }
}
let x = new Foo();
// Expected error here
x.bar = 10;

This is now an error in TypeScript 2.0 and later. See #12 for the suggestion tracking this issue.

Why are function parameters bivariant?

I wrote some code like this and expected an error:

function trainDog(d: Dog) { ... }
function cloneAnimal(source: Animal, done: (result: Animal) => void): void { ... }
let c = new Cat();

// Runtime error here occurs because we end up invoking 'trainDog' with a 'Cat'
cloneAnimal(c, trainDog);

This is an unsoundness resulting from the lack of explicit covariant/contravariant annotations in the type system. Because of this omission, TypeScript must be more permissive when asked whether (x: Dog) => void is assignable to (x: Animal) => void.

To understand why, consider two questions: Is Dog[] a subtype of Animal[] ? Should Dog[] be a subtype of Animal[] in TypeScript?

The second question (should Dog[] be a subtype of Animal[]?) is an easier to analyze. What if the answer was "no" ?

function checkIfAnimalsAreAwake(arr: Animal[]) { ... }

let myPets: Dog[] = [spot, fido];

// Error? Can't substitute Dog[] for Animal[] ?
checkIfAnimalsAreAwake(myPets);

This would be incredibly annoying. The code here is 100% correct provided that checkIfAnimalsAreAwake doesn't modify the array. There's not a good reason to reject this program on the basis that Dog[] can't be used in place of Animal[] - clearly a group of Dogs is a group of Animals here.

Back to the first question. When the type system decides whether or not Dog[] is a subtype of Animal[], it does the following computation (written here as if the compiler took no optimizations), among many others:

  • Is Dog[] assignable to Animal[] ?
  • Is each member of Dog[] assignable to Animal[] ?
    • Is Dog[].push assignable to Animal[].push?
      • Is the type (x: Dog) => number assignable to (x: Animal) => number ?
        • Is the first parameter type in (x: Dog) => number assignable to or from first parameter type in (x: Animal) => number?
          • Is Dog assignable to or from Animal?
            • Yes.

As you can see here, the type system must ask "Is the type (x: Dog) => number assignable to (x: Animal) => number?", which is the same question the type system needed to ask for the original question. If TypeScript forced contravariance on parameters (requiring Animal being assignable to Dog), then Dog[] would not be assignable to Animal[].

In summary, in the TypeScript type system, the question of whether a more-specific-type-accepting function should be assignable to a function accepting a less-specific type provides a prerequisite answer to whether an array of that more specific type should be assignable to an array of a less specific type. Having the latter not be the case would not be an acceptable type system in the vast majority of cases, so we have to take a correctness trade-off for the specific case of function argument types.

Why are functions with fewer parameters assignable to functions that take more parameters?

I wrote some code like this and expected an error:

function handler(arg: string) {
    // ....
}

function doSomething(callback: (arg1: string, arg2: number) => void) {
    callback('hello', 42);
}

// Expected error because 'doSomething' wants a callback of
// 2 parameters, but 'handler' only accepts 1
doSomething(handler);

This is the expected and desired behavior. First, refer to the "substitutability" primer at the top of the FAQ -- handler is a valid argument for callback because it can safely ignored extra parameters.

Second, let's consider another case:

let items = [1, 2, 3];
items.forEach(arg => console.log(arg));

This is isomorphic to the example that "wanted" an error. At runtime, forEach invokes the given callback with three arguments (value, index, array), but most of the time the callback only uses one or two of the arguments. This is a very common JavaScript pattern and it would be burdensome to have to explicitly declare unused parameters.

But forEach should just mark its parameters as optional! e.g. forEach(callback: (element?: T, index?: number, array?: T[]))

This is not what an optional callback parameter means. Function signatures are always read from the caller's perspective. If forEach declared that its callback parameters were optional, the meaning of that is "forEach might call the callback with 0 arguments".

The meaning of an optional callback parameter is this:

// Invoke the provided function with 0 or 1 argument
function maybeCallWithArg(callback: (x?: number) => void) {
    if (Math.random() > 0.5) {
       callback();
    } else {
       callback(42);
    }
}

forEach always provides all three arguments to its callback. You don't have to check for the index argument to be undefined - it's always there; it's not optional.

There is currently not a way in TypeScript to indicate that a callback parameter must be present. Note that this sort of enforcement wouldn't ever directly fix a bug. In other words, in a hypothetical world where forEach callbacks were required to accept a minimum of one argument, you'd have this code:

[1, 2, 3].forEach(() => console.log("just counting"));
             //   ~~ Error, not enough arguments?

which would be "fixed", but not made any more correct, by adding a parameter:

[1, 2, 3].forEach(x => console.log("just counting"));
               // OK, but doesn't do anything different at all

Why are functions returning non-void assignable to function returning void?

I wrote some code like this and expected an error:

function doSomething(): number {
    return 42;
}

function callMeMaybe(callback: () => void) {
    callback();
}

// Expected an error because 'doSomething' returns number, but 'callMeMaybe'
// expects void-returning function
callMeMaybe(doSomething);

This is the expected and desired behavior. First, refer to the "substitutability" primer -- the fact that doSomething returns "more" information than callMeMaybe is a valid substitution.

Second, let's consider another case:

let items = [1, 2];
callMeMaybe(() => items.push(3));

This is isomorphic to the example that "wanted" an error. Array#push returns a number (the new length of the array), but it's a safe substitute to use for a void-returning function.

Another way to think of this is that a void-returning callback type says "I'm not going to look at your return value, if one exists".

Why are all types assignable to empty interfaces?

I wrote some code like this and expected an error:

interface Thing { /* nothing here */ }
function doSomething(a: Thing) {
  // mysterious implementation here
}
// Expected some or all of these to be errors
doSomething(window);
doSomething(42);
doSomething('huh?');

Types with no members can be substituted by any type. In this example, window, 42, and 'huh?' all have the required members of a Thing (there are none).

In general, you should never find yourself declaring an interface with no properties.

Can I make a type alias nominal?

I wrote the following code and expected an error:

type SomeUrl = string;
type FirstName = string;
let x: SomeUrl = "http://www.typescriptlang.org/";
let y: FirstName = "Bob";
x = y; // Expected error

Type aliases are simply aliases -- they are indistinguishable from the types they refer to.

A workaround involving intersection types to make "branded primitives" is possible:

// Strings here are arbitrary, but must be distinct
type SomeUrl = string & {'this is a url': {}};
type FirstName = string & {'person name': {}};

// Add type assertions
let x = <SomeUrl>'';
let y = <FirstName>'bob';
x = y; // Error

// OK
let xs: string = x;
let ys: string = y;
xs = ys;

You'll need to add a type assertion wherever a value of this type is created. These can still be aliased by string and lose type safety.

How do I prevent two types from being structurally compatible?

I would like the following code to produce an error:

interface ScreenCoordinate {
  x: number;
  y: number;
}
interface PrintCoordinate {
  x: number;
  y: number;
}
function sendToPrinter(pt: PrintCoordinate) {
  // ...
}
function getCursorPos(): ScreenCoordinate {
  // Not a real implementation
  return { x: 0, y: 0 };
}
// This should be an error
sendToPrinter(getCursorPos());

A possible fix if you really want two types to be incompatible is to add a 'brand' member:

interface ScreenCoordinate {
  _screenCoordBrand: any;
  x: number;
  y: number;
}
interface PrintCoordinate {
  _printCoordBrand: any;
  x: number;
  y: number;
}

// Error
sendToPrinter(getCursorPos());

Note that this will require a type assertion wherever 'branded' objects are created:

function getCursorPos(): ScreenCoordinate {
  // Not a real implementation
  return <ScreenCoordinate>{ x: 0, y: 0 };
}

See also #202 for a suggestion tracking making this more intuitive.

How do I check at run-time if an object implements some interface?

I want to write some code like this:

interface SomeInterface {
  name: string;
  length: number;
}
interface SomeOtherInterface {
  questions: string[];
}

function f(x: SomeInterface|SomeOtherInterface) {
  // Can't use instanceof on interface, help?
  if (x instanceof SomeInterface) {
    // ...
  }
}

TypeScript types are erased (https://en.wikipedia.org/wiki/Type_erasure) during compilation. This means there is no built-in mechanism for performing runtime type checks. It's up to you to decide how you want to distinguish objects. A popular method is to check for properties on an object. You can use user-defined type guards to accomplish this:

function isSomeInterface(x: any): x is SomeInterface {
  return typeof x.name === 'string' && typeof x.length === 'number';

function f(x: SomeInterface|SomeOtherInterface) {
  if (isSomeInterface(x)) {
    console.log(x.name); // Cool!
  }
}

Why doesn't this incorrect cast throw a runtime error?

I wrote some code like this:

let x: any = true;
let y = <string>x; // Expected: runtime error (can't convert boolean to string)

or this

let a: any = 'hmm';
let b = a as HTMLElement; // expected b === null

TypeScript has type assertions, not type casts. The intent of <T>x is to say "TypeScript, please treat x as a T", not to perform a type-safe run-time conversion. Because types are erased, there is no direct equivalent of C#'s expr as type or (type)expr syntax.

Why don't I get type checking for (number) => string or (T) => T?

I wrote some code like this and expected an error:

let myFunc: (number) => string = (n) => 'The number in hex is ' + n.toString(16);
// Expected error because boolean is not number
console.log(myFunc(true));

Parameter names in function types are required. The code as written describes a function taking one parameter named number of type any. In other words, this declaration

let myFunc: (number) => string;

is equivalent to this one

let myFunc: (number: any) => string;

You should instead write:

let myFunc: (myArgName: number) => string;

To avoid this problem, turn on the noImplicitAny flag, which will issue a warning about the implicit any parameter type.

Why am I getting an error about a missing index signature?

These three functions seem to do the same thing, but the last one is an error. Why is this the case?

interface StringMap {
  [key: string]: string;
}

function a(): StringMap {
  return { a: "1" }; // OK
}

function b(): StringMap {
  var result: StringMap = { a: "1" };
  return result; // OK
}

function c(): StringMap {
  var result = { a: "1" };
  return result; // Error - result lacks index signature, why?
}

This isn't now an error in TypeScript 1.8 and later. As for earlier versions:

Contextual typing occurs when the context of an expression gives a hint about what its type might be. For example, in this initialization:

var x: number = y;

The expression y gets a contextual type of number because it's initializing a value of that type. In this case, nothing special happens, but in other cases more interesting things will occur.

One of the most useful cases is functions:

// Error: string does not contain a function called 'ToUpper'
var x: (n: string) => void = (s) => console.log(s.ToUpper());

How did the compiler know that s was a string? If you wrote that function expression by itself, s would be of type any and there wouldn't be any error issued. But because the function was contextually typed by the type of x, the parameter s acquired the type string. Very useful!

At the same time, an index signature specifies the type when an object is indexed by a string or a number. Naturally, these signatures are part of type checking:

var x: { [n: string]: Car; };
var y: { [n: string]: Animal; };
x = y; // Error: Cars are not Animals, this is invalid

The lack of an index signature is also important:

var x: { [n: string]: Car; };
var y: { name: Car; };
x = y; // Error: y doesn't have an index signature that returns a Car

The problem with assuming that objects don't have index signatures is that you then have no way to initialize an object with an index signature:

var c: Car;
// Error, or not?
var x: { [n: string]: Car } = { 'mine': c };

The solution is that when an object literal is contextually typed by a type with an index signature, that index signature is added to the type of the object literal if it matches. For example:

var c: Car;
var a: Animal;
// OK
var x: { [n: string]: Car } = { 'mine': c };
// Not OK: Animal is not Car
var y: { [n: string]: Car } = { 'mine': a };

Let's look at the original function:

function c(): StringMap {
  var result = { a: "1" };
  return result; // Error - result lacks index signature, why?
}

Because result's type does not have an index signature, the compiler throws an error.

Why am I getting Supplied parameters do not match any signature error?

A function or a method implementation signature is not part of the overloads.

function createLog(message:string): number;
function createLog(source:string, message?:string): number {
  return 0;
}

createLog("message"); // OK
createLog("source", "message"); // ERROR: Supplied parameters do not match any signature

When having at least one overload signature declaration, only the overloads are visible. The last signature declaration, also known as the implementation signature, does not contribute to the shape of your signature. So to get the desired behavior you will need to add an additional overload:

function createLog(message:string): number;
function createLog(source:string, message:string): number
function createLog(source:string, message?:string): number {
  return 0;
}

The rationale here is that since JavaScript does not have function overloading, you will be doing parameter checking in your function, and this your function implementation might be more permissive that what you would want your users to call you through.

For instance you can require your users to call you using matching pairs of arguments, and implement this correctly without having to allow mixed argument types:

function compare(a: string, b: string): void;
function compare(a: number, b: number): void;
function compare(a: string|number, b: string|number): void {
  // Just an implementation and not visible to callers
}

compare(1,2) // OK
compare("s", "l") // OK
compare (1, "l") // Error.

Functions

Why can't I use x in the destructuring function f({ x: number }) { /* ... */ }?

I wrote some code like this and got an unexpected error:

function f({x: number}) {
  // Error, x is not defined?
  console.log(x);
}

Destructuring syntax is counterintuitive for those accustomed to looking at TypeScript type literals. The syntax f({x: number}) declares a destructuring from the property x to the local number.

Looking at the emitted code for this is instructive:

function f(_a) {
    // Not really what we were going for
    var number = _a.x;
}

To write this code correctly, you should write:

function f({x}: {x: number}) {
  // OK
  console.log(x);
}

If you can provide a default for all properties, it's preferable to write:

function f({x = 0}) {
  // x: number
  console.log(x);
}

Classes

Why do these empty classes behave strangely?

I wrote some code like this and expected an error:

class Empty { /* empty */ }

var e2: Empty = window;

See the question "Why are all types assignable to empty interfaces?" in this FAQ. It's worth re-iterating the advice from that answer: in general, you should never declare a class with no properties. This is true even for subclasses:

class Base {
  important: number;
  properties: number;
}
class Alpha extends Base { }
class Bravo extends Base { }

Alpha and Bravo are structurally identical to each other, and to Base. This has a lot of surprising effects, so don't do it! If you want Alpha and Bravo to be different, add a private property to each.

When and why are classes nominal?

What explains the difference between these two lines of code?

class Alpha { x: number }
class Bravo { x: number }
class Charlie { private x: number }
class Delta { private x: number }

let a = new Alpha(), b = new Bravo(), c = new Charlie(), d = new Delta();

a = b; // OK
c = d; // Error

In TypeScript, classes are compared structurally. The one exception to this is private and protected members. When a member is private or protected, it must originate in the same declaration to be considered the same as another private or protected member.

Why does this get orphaned in my instance methods?

I wrote some code like this:

class MyClass {
  x = 10;
  someCallback() {
    console.log(this.x); // Prints 'undefined', not 10
    this.someMethod(); // Throws error "this.method is not a function"
  }
  someMethod() {

  }
}

let obj = new MyClass();
window.setTimeout(obj.someCallback, 10);

Synonyms and alternate symptoms:

  • Why are my class properties undefined in my callback?
  • Why does this point to window in my callback?
  • Why does this point to undefined in my callback?
  • Why am I getting an error this.someMethod is not a function?
  • Why am I getting an error Cannot read property 'someMethod' of undefined ?

In JavaScript, the value of this inside a function is determined as follows:

  1. Was the function the result of calling .bind? If so, this is the first argument passed to bind
  2. Was the function directly invoked via a property access expression expr.method() ? If so, this is expr
  3. Otherwise, this is undefined (in "strict" mode), or window in non-strict mode

The offending problem is this line of code:

window.setTimeout(obj.someCallback, 10);

Here, we provided a function reference to obj.someCallback to setTimeout. The function was then invoked on something that wasn't the result of bind and wasn't directly invoked as a method. Thus, this in the body of someCallback referred to window (or undefined in strict mode).

Solutions to this are outlined here: http://stackoverflow.com/a/20627988/1704166

What's the difference between Bar and typeof Bar when Bar is a class ?

I wrote some code like this and don't understand the error I'm getting:

class MyClass {
  someMethod() { }
}
var x: MyClass;
// Cannot assign 'typeof MyClass' to MyClass? Huh?
x = MyClass;

It's important to remember that in JavaScript, classes are just functions. We refer to the class object itself -- the value MyClass -- as a constructor function. When a constructor function is invoked with new, we get back an object that is an instance of the class.

So when we define a class, we actually define two different types.

The first is the one referred to by the class's name; in this case, MyClass. This is the instance type of the class. It defines the properties and methods that an instance of the class has. It's the type returned by invoking the class's constructor.

The second type is anonymous. It is the type that the constructor function has. It contains a construct signature (the ability to be invoked with new) that returns an instance of the class. It also contains any static properties and methods the class might have. This type is typically referred to as the "static side" of the class because it contains those static members (as well as being the constructor for the class). We can refer to this type with the type query operator typeof.

The typeof operator (when used in a type position) expresses the type of an expression. Thus, typeof MyClass refers to the type of the expression MyClass - the constructor function that produces instances of MyClass.

Why do my derived class property initializers overwrite values set in the base class constructor?

See #1617 for this and other initialization order questions

What's the difference between declare class and interface?

TODO: Write up common symptoms of declare class / interface confusion.

See http://stackoverflow.com/a/14348084/1704166

What does it mean for an interface to extend a class?

What does this code mean?

class Foo {
  /* ... */
}
interface Bar extends Foo {

}

This makes a type called Bar that has the same members as the instance shape of Foo. However, if Foo has private members, their corresponding properties in Bar must be implemented by a class which has Foo in its heritage. In general, this pattern is best avoided, especially if Foo has private members.

Why am I getting "TypeError: [base class name] is not defined in __extends ?

I wrote some code like this:

/** file1.ts **/
class Alpha { /* ... */ }

/** file2.ts **/
class Bravo extends Alpha { /* ... */ }

I'm seeing a runtime error in __extends:

Uncaught TypeError: Alpha is not defined

The most common cause of this is that your HTML page includes a <script> tag for file2.ts, but not file1.ts. Add a script tag for the base class's output before the script tag for the derived class.

Why am I getting "TypeError: Cannot read property 'prototype' of undefined" in __extends ?

I wrote some code:

/** file1.ts **/
class Alpha { /* ... */ }

/** file2.ts **/
class Bravo extends Alpha { /* ... */ }

I'm seeing a runtime error in __extends:

Uncaught TypeError: Cannot read property 'prototype' of undefined

This can happen for a few reasons.

The first is that, within a single file, you defined the derived class before the base class. Re-order the file so that base classes are declared before the derived classes.

If you're using --out, the compiler may be confused about what order you intended the files to be in. See the section "How do I control file ordering..." in the FAQ.

If you're not using --out, your script tags in the HTML file may be the wrong order. Re-order your script tags so that files defining base classes are included before the files defining the derived classes.

Finally, if you're using a third-party bundler of some sort, that bundler may be ordering files incorrectly. Refer to that tool's documentation to understand how to properly order the input files in the resulting output.

Why doesn't extending built-ins like Error, Array, and Map work?

In ES2015, constructors which return an object implicitly substitute the value of this for any callers of super(...). It is necessary for generated constructor code to capture any potential return value of super(...) and replace it with this.

As a result, subclassing Error, Array, and others may no longer work as expected. This is due to the fact that constructor functions for Error, Array, and the like use ECMAScript 6's new.target to adjust the prototype chain; however, there is no way to ensure a value for new.target when invoking a constructor in ECMAScript 5. Other downlevel compilers generally have the same limitation by default.

Example

For a subclass like the following:

class FooError extends Error {
    constructor(m: string) {
        super(m);
    }
    sayHello() {
        return "hello " + this.message;
    }
}

you may find that:

  • methods may be undefined on objects returned by constructing these subclasses, so calling sayHello will result in an error.
  • instanceof will be broken between instances of the subclass and their instances, so (new FooError()) instanceof FooError will return false.

Recommendation

As a recommendation, you can manually adjust the prototype immediately after any super(...) calls.

class FooError extends Error {
    constructor(m: string) {
        super(m);

        // Set the prototype explicitly.
        Object.setPrototypeOf(this, FooError.prototype);
    }

    sayHello() {
        return "hello " + this.message;
    }
}

However, any subclass of FooError will have to manually set the prototype as well. For runtimes that don't support Object.setPrototypeOf, you may instead be able to use __proto__.

Unfortunately, these workarounds will not work on Internet Explorer 10 and prior. One can manually copy methods from the prototype onto the instance itself (i.e. FooError.prototype onto this), but the prototype chain itself cannot be fixed.

Generics

Why is A<string> assignable to A<number> for interface A<T> { }?

I wrote some code and expected an error:

interface Something<T> {
  name: string;
}
let x: Something<number>;
let y: Something<string>;
// Expected error: Can't convert Something<number> to Something<string>!
x = y;

TypeScript uses a structural type system. When determining compatibility between Something<number> and Something<string>, we examine each member of each type. If each member of the types are compatible, then the type are compatible as well. Because Something<T> doesn't use T in any member, it doesn't matter what type T is.

In general, you should never have a type parameter which is unused. The type will have unexpected compatibility (as shown here) and will also fail to have proper generic type inference in function calls.

Why doesn't type inference work on this interface: interface Foo<T> { } ?

I wrote some code like this:

interface Named<T> {
    name: string;
}
class MyNamed<T> implements Named<T> {
  name: 'mine';
}
function findByName<T>(x: Named<T>): T {
  // TODO: Implement
  return undefined;
}

var x: MyNamed<string>;
var y = findByName(x); // expected y: string, got y: {}

TypeScript uses a structural type system. This structuralness also applies during generic type inference. When inferring the type of T in the function call, we try to find members of type T on the x argument to figure out what T should be. Because there are no members which use T, there is nothing to infer from, so we return {}.

Note that if you use T, you get correct inference:

interface Named<T> {
    name: string;
  value: T; // <-- added
}
class MyNamed<T> implements Named<T> {
  name: 'mine';
  value: T; // <-- added
}
function findByName<T>(x: Named<T>): T {
  // TODO: Implement
  return undefined;
}

var x: MyNamed<string>;
var y = findByName(x); // got y: string;

Remember: You should never have unused type parameters! See the previous question for more reasons why this is bad.

Why can't I write typeof T, new T, or instanceof T in my generic function?

I want to write some code like this:

function doSomething<T>(x: T) {
  // Can't find name T?
  let xType = typeof T;
  let y = new xType();
  // Same here?
  if(someVar instanceof typeof T) {

  }
  // How do I instantiate?
  let z = new T();
}

Generics are erased during compilation. This means that there is no value T at runtime inside doSomething. The normal pattern that people try to express here is to use the constructor function for a class either as a factory or as a runtime type check. In both cases, using a construct signature and providing it as a parameter will do the right thing:

function create<T>(ctor: { new(): T }) {
    return new ctor();
}
var c = create(MyClass); // c: MyClass

function isReallyInstanceOf<T>(ctor: { new(...args: any[]): T }, obj: T) {
  return obj instanceof ctor;
}

Modules

Why are imports being elided in my emit?

I wrote some code like this

import someModule = require('./myMod');

let x: someModule.SomeType = /* something */;

and the emit looked like this:

// Expected to see "var someModule = require('./myMod');" here!

var x = /* something */;

TypeScript assumes that module imports do not have side effects, so it removes module imports that aren't used in any expression.

Use import "mod" syntax to force the module to be loaded.

import "./myMod"; // For side effects

You can also simply reference the module. This is the most universal workaround. A single use will do:

import someModule = require('./myMod');
someModule; // Used for side effects

Why don't namespaces merge across different module files?

TODO: Port content from http://stackoverflow.com/questions/30357634/how-do-i-use-namespaces-with-typescript-external-modules

Enums

What's the difference between enum and const enums?

TODO: Write up common symptoms of enum / const enum confusion.

See http://stackoverflow.com/questions/28818849/how-do-the-different-enum-variants-work-in-typescript

Type Guards

Why doesn't x instanceof Foo narrow x to Foo?

It depends what x is. If the type of x was originally not even compatible with Foo, then it wouldn't make much sense to narrow the type, so we don't.

More likely, you'll find yourself in this situation when x had the type any. The motivating example for this is something like the following:

function doIt(x) {
    if (x instanceof Object) {
        // Assume 'x' is a well-known object which
        // we know how to handle specifically
    }

    // Treat 'x' as a primitive
}

You'll see this type of code in TypeScript code that predates union types, or TypeScript code that's been ported over from JavaScript. If we narrowed from any to Object then there's not much you could really do with x. Using any properties that aren't in Object will lead to an error. This is not just true of Object, it's true of any other type with a defined set of properties.

Why doesn't isFoo(x) narrow x to Foo when isFoo is a type guard?

TODO, but it is strongly related to the above section.

Decorators

Decorators on function declarations

TODO: Answer. Also, what did we mean here?

What's the difference between @dec and @dec() ? Shouldn't they be equivalent?

TODO: Answer

JSX and React

I wrote declare var MyComponent: React.Component;, why can't I write <MyComponent /> ?

I wrote some code like this. Why is there an error?

class Display extends React.Component<any, any> {
    render() { ... }
}

let SomeThing: Display = /* ... */;
// Error here, isn't this OK?
let jsx = <SomeThing />;

This is a confusion between the instance and static side of a class. When React instantiates a component, it's invoking a constructor function. So when TypeScript sees a JSX <TagName />, it is validating that the result of constructing TagName produces a valid component.

But by declaring let SomeThing: Display, the code is indicating that SomeThing is the class instance, not the class constructor. Indeed, it would be a run-time error to write:

let SomeThing = new Display();
let jsx = <SomeThing />; // Not gonna work

The easiest fix is to use the typeof type operator.

let SomeThing: typeof Display = /* ... */;

Things That Don't Work

You should emit classes like this so they have real private members

If I write code like this:

class Foo {
    private x = 0;
    increment(): number {
        this.x++;
        return x;
    }
}

You should emit code like this so that 'x' is truly private:

var Foo = (function () {
    var x = 0;

    function Foo() {
    }
    Foo.prototype.increment = function () {
        x++;
        return x;
    };
    return Foo;
})();

This code doesn't work. It creates a single private field that all classes share:

var a = new Foo();
a.increment(); // Prints 1
a.increment(); // Prints 2
var b = new Foo(); // Should not affect a
a.increment(); // Prints 1

You should emit classes like this so they don't lose this in callbacks

If I write code like this:

class MyClass {
	method() {
	}
}

You should emit code like this so that I can't mess up this in in callbacks:

var MyClass = (function () {
    function MyClass() {
		this.method = function() {

		}
    }
    return MyClass;
})();

Two problems here.

First, the proposed behavior change is not in line with the ECMAScript specification. There isn't really anything else to be said on that front -- TypeScript must have the same runtime behavior as JavaScript.

Second, the runtime characteristics of this class are very surprising. Instead of allocating one closure per method, this allocates one closure per method per instance. This expensive in terms of class initialization cost, memory pressure, and GC performance.

Why can't I access arbitrary properties on a type with a string indexer?

When I declare an interface with a string indexer, I want to be able to access arbitrary properties on it, like this:

interface StringMap {
   [index: string]: string;
}
function f(obj: StringMap) {
   obj.foo; // error!
   obj['foo']; // I have to write it this way
}

Instead I can only use index syntax: obj['foo'].

The point of TypeScript is to catch errors at compile-time, and "Property does not exist" is one of the most important. If a string indexer let you access properties on a type, you'd never get this error for those types. This is important if you have other properties on the indexed type:

interface StringMap {
   property1: string;
   [index: string]: string;
}
function f(obj: StringMap) {
   obj.property1; // ok
   obj.propertyl; // error!
   obj['foo']; // ok
}

So TypeScript always checks property accesses and reserves arbitrary access for the indexer syntax.

You should have some class initialization which is impossible to emit code for

TODO: Port content from #1617

External Tools

How do I write unit tests with TypeScript?

Commandline Behavior

Why did adding an import or export modifier break my program?

I wrote a program:

/* myApp.ts */
function doSomething() {
    console.log('Hello, world!');
}
doSomething();

I compiled it with tsc --module commonjs myApp.ts --out app.js and ran node app.js and got the expected output.

Then I added an import to it:

import fs = require('fs');
function doSomething() {
    console.log('Hello, world!');
}
doSomething();

Or added an export to it:

export function doSomething() {
    console.log('Hello, world!');
}
doSomething();

And now nothing happens when I run app.js!

Modules -- those files containing top-level export or import statements -- are always compiled 1:1 with their corresponding js files. The --out option only controls where script (non-module) code is emitted. In this case, you should be running node myApp.js, because the module myApp.ts is always emitted to the file myApp.js.

This behavior has been fixed as of TypeScript 1.8; combining --out and --module is now an error for CommonJS module output.

How do I control file ordering in combined output (--out) ?

The order of the generated files in the output follows that of the input files after the pre-processing pass.

The compiler performs a pre-processing pass on input files to resolve all triple-slash reference directives and module import statements. During this process, additional files can be added to the compilation.

The process starts with a set of root files; these are the file names specified on the command-line or in the "files" list in the tsconfig.json file. These root files are pre-processed in the same order they are specified. Before a file is added to the list, all triple-slash references and import statements in it are processed, and their targets included. Triple-slash references and import statements are resolved in a depth-first manner, in the order they appear in the file.

See more information about resolving triple-slash reference directives at triple-slash directives documentation and module import statements resolution at module resolution documentation.

What does the error "Exported variable [name] has or is using private name [name]" mean?

This error occurs when you use the --declaration flag because the compiler is trying to produce a declaration file that exactly matches the module you defined.

Let's say you have this code:

/// MyFile.ts
class Test {
    // ... other members ....
    constructor(public parent: Test){}
}

export let t = new Test("some thing");

To produce a declaration file, the compiler has to write out a type for t:

/// MyFile.d.ts, auto-generated
export let t: ___fill in the blank___;

The member t has the type Test. The type Test is not visible because it's not exported, so we can't write t: Test.

In the very simplest cases, we could rewrite Test's shape as an object type literal. But for the vast majority of cases, this doesn't work. As written, Test's shape is self-referential and can't be written as an anonymous type. This also doesn't work if Test has any private or protected members. So rather than let you get 65% of the way through writing a realistic class and then start erroring then, we just issue the error (you're almost certainly going to hit later anyway) right away and save you the trouble.

To avoid this error:

  1. Export the declarations used in the type in question
  2. Specify an explicit type annotation for the compiler to use when writing declarations.

Why does --outDir moves output after adding a new file?

--outDir specifies the "root" directory of the output. The compiler needs a "root" directory in the source to mirror into the output directory. If --rootDir is not specified, the compiler will compute one; this is based on a common path calculation, which is the longest common prefix of all your input files. Obviously this changes with adding a new file to the compilation that has a shorter path prefix.

To ensure the output does not change with adding new files specify --rootDir on the command-line or in your tsconfig.json.

tsconfig.json Behavior

Why is a file in the exclude list still picked up by the compiler?

tsconfig.json turns a folder into a “project”. Without specifying any “exclude” or “files” entries, all files in the folder containing the tsconfig.json and all its sub-directories are included in your compilation.

If you want to exclude some of the files use “exclude”, if you would rather specify all the files instead of letting the compiler look them up, use “files”.

That was tsconfig.json automatic inclusion. There is a different issue, which is module resolution. By module resolution, I mean the compiler trying to understand what ns means in an import statement like: import * ns from “mod”. To do so, the compiler needs the definition of a module, this could be a .ts file for your own code, or a .d.ts for an imported definition files. if the file was found, it will be included regardless if it was excluded in the previous steps.

So to exclude a file from the compilation, you need to exclude and all all files that has an import or /// <reference path="..." /> directives to it.

Use tsc --listFiles to list what files are included in your compilation, and tsc --traceResolution to see why they were included.

How can I specify an include?

There is no way now to indicate an “include” to a file outside the current folder in the tsconfig.json (tracked by #1927). You can achieve the same result by either: 1. Using a “files” list, or 2. Adding a /// <reference path="..." /> directive in one of the files in your directory.

Why am I getting the error TS5055: Cannot write file 'xxx.js' because it would overwrite input file. when using JavaScript files?

For a TypeScript file, the TypeScript compiler by default emits the generated JavaScript files in the same directory with the same base file name. Because the TypeScript files and emitted JavaScript files always have different file extensions, it is safe to do so. However, if you have set the allowJs compiler option to true and didn't set any emit output options (outFile and outDir), the compiler will try to emit JavaScript source files by the same rule, which will result in the emitted JavaScript file having the same file name with the source file. To avoid accidentally overwriting your source file, the compiler will issue this warning and skip writing the output files.

There are multiple ways to solve this issue, though all of them involve configuring compiler options, therefore it is recommended that you have a tsconfig.json file in the project root to enable this. If you don't want JavaScript files included in your project at all, simply set the allowJs option to false; If you do want to include and compile these JavaScript files, set the outDir option or outFile option to direct the emitted files elsewhere, so they won't conflict with your source files; If you just want to include the JavaScript files for editing and don't need to compile, set the noEmit compiler option to true to skip the emitting check.

Comments

Why some comments are not preserved in emitted JavaScript even when --removeComments is not specified?

TypeScript compiler uses a position of a node in the abstract syntax tree to retrieve its comments during emit. Because, the compiler does not store all tokens into the tree, some comments may be missed in an output JavaScript file. For example, we do not store following tokens into the tree ,, {, }, (, ). Therefore, trailing comments or leading comments of such token cannot be retrieved during emit. At the moment, there is not an easy method to preserve such comments without storing those tokens. Doing so, however, can significantly increase the tree size and potentially have performance impact.

Some cases where TypeScript compiler will not be able to preserve your comments:

/* comment */
<div>
    {/* comment will not be emitted */}
</div>

var x = {
    prop1: 1, // won't get emit because we can't retrieve this comment
    prop2: 2  // will be emit
}

function foo() /* this comment can't be preserved */ { }

Why Copyright comments are removed when --removeComments is true?

TypeScript compiler will preserve copyright comment regardless of --removeComments. For a comment to be considered a copyright comment, it must have following characteristics:

  • a top-of-file comment following by empty line, separating it from the first statement.
  • begin with /*!

Glossary and Terms in this FAQ

Dogs, Cats, and Animals, Oh My

For some code examples, we'll use a hypothetical type hierarchy:

       Animal
      /      \
    Dog      Cat

That is, all Dogs are Animals, all Cats are Animals, but e.g. a function expecting a Dog cannot accept an argument of type Cat. If you want to try these examples in the TypeScript Playground, start with this template:

interface Animal {
  move(): void;
}
interface Dog extends Animal {
  woof: string;
}
interface Cat extends Animal {
  meow: string;
}

Other examples will use DOM types like HTMLElement and HTMLDivElement to highlight concrete real-world implications of certain behaviors.

"Substitutability"

Many answers relating to the type system make reference to Substitutability. This is a principle that says if an object X can be used in place of some object Y, then X is a subtype of Y. We also commonly say that X is assignable to Y (these terms have slightly different meanings in TypeScript, but the difference is not important here).

In other words, if I ask for a fork, a spork is an acceptable substitute because it has the same functions and properties of a fork (three prongs and a handle).

Trailing, leading, and detached comments

TypeScript classifies comments into three different types:

  • Leading comment : a comment before a node followed by newline.
  • Trailing comment : a comment after a node and in the same line as the node.
  • Detached comment : a comment that is not part of any node such as copyright comment.
/*! Top-of-file copyright comment is a detached comment*/

/* Leading comments of the function AST node*/
function foo /*trailing comments of the function name, "foo", AST node*/ () {
  /* Detached comment*/

  let x = 10;
}

GitHub Process Questions

What do the labels on these issues mean?

What are all the labels people keep putting on my issues?

  • help wanted: We are accepting pull requests to implement this feature or fix this bug. PRs must adhere to the rules specified in CONTRIBUTING.md
  • Breaking Change: Fixing this bug or implementing this feature will break code that someone could have plausibly written (i.e. we do not consider new errors in nonsense code like undefined.throwSomething() to be breaking changes)
  • By Design: This is an intentional behavior of TypeScript
  • Canonical: This issue contains a lengthy explanation of a common question or misconception
  • Committed: Someone from the TypeScript team will fix this bug or implement this feature
  • Declined: For reasons explained in the issue, we are not going to accept this suggestion (note: See "I disagree with the outcome..." section)
  • Discussion: This issue is a discussion with no defined outcome. The TypeScript team may weigh in on these issues, but they are not regularly reviewed
  • Duplicate: This issue is the same, or has the same root cause, as another issue
  • Effort: Easy/Moderate/Difficult: For issues marked as 'help wanted', these are an approximation of how difficult we think fixing the bug or implementing the feature will be. As a rough guide, fixing typos or modifying lib.d.s are generally Easy; work that requires understanding the basics of the code base is Moderate; things marked Difficult will require an understanding that is rare outside the core TypeScript team
  • good first issue: These are 'Effort: easy' issues, good for your first contribution
  • ES6 / ES7 / ES Next: Refers to issues related to features found in these specific ECMAScript versions
  • External: Catch-all bucket when an issue reported is not an issue with TypeScript, but rather an external tool, library, website, person, or situation
  • Fixed: This bug has been fixed. Generally, you will see these bugs fixed in the nightly version(npm install typescript@next) within 24-48 hours
  • High Priority: Issues affecting runtime behavior or high-occurrence crashes
  • Infrastructure: Technical debt associated with the TypeScript project
  • In Discussion: The suggestion is ready to be discussed at a Design Meeting or Suggestion Backlog Slog
  • Needs More Info: The team needs more information about this suggestion or bug in order to understand what's going on. Generally, Suggestions will start out as Needs More Info, graduate to Needs Proposal, then finally go to In Discussion
  • Needs Proposal: A suggestion that has a well-understood use case and a plausible outline of a solution, but lacks a formal definition of how exactly the problem will be solved
  • Out of Scope: A suggestion that is outside the design parameters of TypeScript, either because it is a poor fit (e.g. make TypeScript look exactly like C#), is outside the constraints of the language (e.g. asm.js compilation), or better belongs to another tool or process (e.g. a built-in Collections library, or a runtime language feature that should start in the ECMAScript committee)
  • Question: The issue is (intentionally or otherwise) simply asking a question about TypeScript. Answers to Questions, if provided, will generally be to-the-point because we do not have time to be a support community for all TypeScript users; please use Stack Overflow for TypeScript questions.
  • Revisit: A suggestion or bug that can't be adequately addressed today, but will probably be able to be addressed in the future (e.g. we need to wait for the ECMAScript committee to make up its mind)
  • Suggestion: Any suggestion
  • Too Complex: Relative to the complexity required to implement or understand it, the suggestion does not provide enough value. This is a subjective measure, see "I disagree with the outcome...")
  • Won't Fix: While the behavior described is agreed to be incorrect, the cost (in time, complexity, performance, etc.) is too high to justify taking a fix relative to the cost of simply living with the bug

I disagree with the outcome of this suggestion

I don't think this suggestion should have been closed! What can I do next?

To date, we've received over 1,000 suggestions on the TypeScript GitHub repo. We do our best to read, understand, prioritize, formalize, and implement these suggestions. User feedback has been critical in shaping the success of the project. That said, sometimes we'll make decisions that you don't agree with, and sometimes we'll make the wrong call. What should you do if you think we should reconsider something? Let's walk through the five stages of grief.

Denial: It's healthy to believe that a suggestion might come back later. Do keep leaving feedback! We look at all comments on all issues - closed or otherwise. If you encounter a problem that would have been addressed by a suggestion, leave a comment explaining what you were doing and how you could have had a better experience. Having a record of these use cases helps us reprioritize.

Anger: Don't be angry. Specifically, remember that the TypeScript team does not have the resources to continuously relitigate closed suggestions.

Bargaining: Ask yourself: is there a smaller thing that would work? Many suggestions are simply too large of a hammer for too small of a nail. Think about the problem you're experiencing for a while and see if you can come up with a simpler solution that accomplishes the same goal.

Depression: Try not to be depressed about declined suggestions. The features that make it into the language instead might solve your problem in an even better way than you could have imagined.

Acceptance: Repeat this mantra: I will accept the features I cannot have, have courage to submit pull requests for those I can, and the wisdom to know the difference by looking at the GitHub labels.

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