Typescript: `!.` after `?.` should be warned

[falsandtru]

This is an anti-pattern on the current type system. And the compiler must not do quick fix so. !. after ?. is simply replaceable with ?. and should do so.

TypeScript Version: 3.7.x-dev.20191105

Search Terms:

Code

document.querySelector('_')!.textContent!.split('') // old code
document.querySelector('_')?.textContent!.split('') ?? 0 // string[]
document.querySelector('_')?.textContent?.split('') ?? 0 // string[] | 0

Expected behavior:

document.querySelector('_')!.textContent!.split('') // old code
document.querySelector('_')?.textContent!.split('') ?? 0 // unsafe, warning
document.querySelector('_')?.textContent?.split('') ?? 0 // safe

Actual behavior:

document.querySelector('_')!.textContent!.split('') // old code
document.querySelector('_')?.textContent!.split('') ?? 0 // unsafe, no warning
document.querySelector('_')?.textContent?.split('') ?? 0 // safe

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Related Issues:

[kitsonk]

Isn't this a just linting rule? Whenever you use ! it is potentially unsafe. The compiler never warns, an error is an error.

[falsandtru]

When quick fix observes this rule, it is wasteful splitting.

[RyanCavanaugh]

@falsandtru some discussion about what exactly is different between this and #35025 would be nice

[falsandtru]

This issue has meaning without adopting #35025 and meaningless if #35025 is adopted. So this issue and #35025 are in opposition. This is the definitive difference.