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Fix incorrect removal of parentheses when using an infer with a constraint in a function predicate #14279

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merged 5 commits into from Feb 6, 2023
Merged

Fix incorrect removal of parentheses when using an infer with a constraint in a function predicate #14279

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86acf97
fix: paren shouldn't be removed from infer func
fisker Jan 31, 2023
ffe71f8
Simplify logic
fisker Jan 31, 2023
467a158
Test `TSConstructorType`
fisker Jan 31, 2023
ac1aac1
Put more stuff inside
fisker Jan 31, 2023
6117e1b
Revert "Put more stuff inside"
fisker Jan 31, 2023
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16 changes: 16 additions & 0 deletions changelog_unreleased/typescript/14279.md
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@@ -0,0 +1,16 @@
#### Fix parens in inferred function return types with `extends` (#14279 by @fisker)

<!-- prettier-ignore -->
```ts
// Input
type Foo<T> = T extends ((a) => a is infer R extends string) ? R : never;

// Prettier stable (First format)
type Foo<T> = T extends (a) => a is infer R extends string ? R : never;

// Prettier stable (Second format)
SyntaxError: '?' expected.

// Prettier main
type Foo<T> = T extends ((a) => a is infer R extends string) ? R : never;
```
24 changes: 16 additions & 8 deletions src/language-js/needs-parens.js
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Expand Up @@ -466,22 +466,30 @@ function needsParens(path, options) {
}

case "TSConditionalType":
if (name === "extendsType" && parent.type === "TSConditionalType") {
return true;
}
// fallthrough
case "TSFunctionType":
case "TSConstructorType":
if (name === "extendsType" && parent.type === "TSConditionalType") {
const returnTypeAnnotation = (node.returnType || node.typeAnnotation)
.typeAnnotation;
if (node.type === "TSConditionalType") {
return true;
}

let { typeAnnotation } = node.returnType || node.typeAnnotation;

if (
typeAnnotation.type === "TSTypePredicate" &&
typeAnnotation.typeAnnotation
) {
typeAnnotation = typeAnnotation.typeAnnotation.typeAnnotation;
}

if (
returnTypeAnnotation.type === "TSInferType" &&
returnTypeAnnotation.typeParameter.constraint
typeAnnotation.type === "TSInferType" &&
typeAnnotation.typeParameter.constraint
) {
return true;
}
}

if (name === "checkType" && parent.type === "TSConditionalType") {
return true;
}
Expand Down
68 changes: 54 additions & 14 deletions tests/format/typescript/conditional-types/__snapshots__/jsfmt.spec.js.snap
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Expand Up @@ -278,20 +278,6 @@ type Unpacked<T> = T extends (infer U)[]
================================================================================
`;

exports[`issue-13275.ts format 1`] = `
====================================options=====================================
parsers: ["typescript"]
printWidth: 80
| printWidth
=====================================input======================================
type Foo<T> = T extends ((...a: any[]) => infer R extends string) ? R : never;

=====================================output=====================================
type Foo<T> = T extends ((...a: any[]) => infer R extends string) ? R : never;

================================================================================
`;

exports[`nested-in-condition.ts format 1`] = `
====================================options=====================================
parsers: ["typescript"]
Expand Down Expand Up @@ -431,3 +417,57 @@ type Unpacked<T> = T extends (infer U)[]

================================================================================
`;

exports[`parentheses.ts format 1`] = `
====================================options=====================================
parsers: ["typescript"]
printWidth: 80
| printWidth
=====================================input======================================
// #13275
type Foo<T> = T extends ((...a: any[]) => infer R extends string) ? R : never;
type Foo<T> = T extends (new (...a: any[]) => infer R extends string) ? R : never;

// #14275
type Test<T> = T extends ((
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
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@fisker fisker Jan 31, 2023 •
edited

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@bradzacher Is it possible to be longer/deeper?

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I don't think it can be any deeper. I'm not 100% certain though. You could probably nest more conditionals in place of the TSESTree.Token - would be worth playing to see what's allowed syntactically.

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It seems possible and I don't know how to fix, I'm going to pretend I don't know. ac1aac1

? U
: TSESTree.Token;
type Test<T> = T extends ((
token: TSESTree.Token
) => asserts token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;
type Test<T> = T extends (new (
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;

=====================================output=====================================
// #13275
type Foo<T> = T extends ((...a: any[]) => infer R extends string) ? R : never;
type Foo<T> = T extends (new (...a: any[]) => infer R extends string)
? R
: never;

// #14275
type Test<T> = T extends ((
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;
type Test<T> = T extends ((
token: TSESTree.Token
) => asserts token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;
type Test<T> = T extends (new (
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;

================================================================================
`;
1 change: 0 additions & 1 deletion tests/format/typescript/conditional-types/issue-13275.ts
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20 changes: 20 additions & 0 deletions tests/format/typescript/conditional-types/parentheses.ts
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@@ -0,0 +1,20 @@
// #13275
type Foo<T> = T extends ((...a: any[]) => infer R extends string) ? R : never;
type Foo<T> = T extends (new (...a: any[]) => infer R extends string) ? R : never;

// #14275
type Test<T> = T extends ((
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;
type Test<T> = T extends ((
token: TSESTree.Token
) => asserts token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;
type Test<T> = T extends (new (
token: TSESTree.Token
) => token is infer U extends TSESTree.Token)
? U
: TSESTree.Token;