vi.unmock does not unmock imitated module

[mcous]

Describe the bug

When mocking modules using vi.mock (or vi.doMock), if the module is replaced implicitly, a subsequent call to vi.unmock (or vi.doUnmock) will not unmock the module.

import {vi, afterEach, test, expect} from 'vitest'

afterEach(() => {
  vi.doUnmock('./dependency')
})

test('mock by imitating the actual module', async () => {
  vi.doMock('./dependency')
  const dependency = await import('./dependency')
  expect(vi.mocked(dependency.doSomething).mock).toBeDefined()
})

test('verify module is not mocked after call to doUnmock', async () => {
  const dependency = await import('./dependency')
  expect(vi.mocked(dependency.doSomething).mock).not.toBeDefined()
})

I believe this to be a bug, because if you provide an explicit replacement via the second argument to vi.mock, instead of relying on the module imitation logic, the call to unmock will successfully remove the mock.

Reproduction

See mcous/vitest-unmock-bug

System Info

System:
    OS: macOS 12.5.1
    CPU: (8) x64 Intel(R) Core(TM) i7-4870HQ CPU @ 2.50GHz
    Memory: 2.06 GB / 16.00 GB
    Shell: 5.9 - /usr/local/bin/zsh
  Binaries:
    Node: 16.18.1 - ~/.nvs/default/bin/node
    npm: 8.19.2 - ~/.nvs/default/bin/npm
    Watchman: 2022.11.07.00 - /usr/local/bin/watchman
  Browsers:
    Firefox: 107.0
    Safari: 15.6.1
  npmPackages:
    vite: ^3.2.4 => 3.2.4
    vitest: ^0.25.2 => 0.25.2

Used Package Manager

pnpm

Validations

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• Read the Contributing Guidelines.
• Read the docs.
• Check that there isn't already an issue that reports the same bug to avoid creating a duplicate.
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• The provided reproduction is a minimal reproducible example of the bug.

[sheremet-va]

This should have „in“ check, because mocks without factory are stored as „null“:

vitest/packages/vitest/src/runtime/mocker.ts

Line 294 in 2499828

if (mock?.[id])

PR welcome

[sheremet-va]

Fixed by #2353