Suggestion: noInferredAny

Search Terms

noImplicitAny inference any noInferredAny

Suggestion

Symbols typed as any reduce the type safety of a program. That's fair if the programmer intended to use any. However it's relatively easy to end up with symbols typed any through type inference chains that are not obvious:

• 3rd party libraries might return any
• complex type expressions might infer to any
• some .d.ts types default generic arguments to any (e.g. Set)

In all these situations, a programmer might write const foo = something(); and expect foo to have a reasonable inferred type. foo being inferred to any is easy to miss in such code, both while editing and while reviewing code.

Proposal: add a compiler option noInferredAny that flags symbols whose type is any and that do not have an explicit type annotation.

Use Cases

• better type safety for programs
• detect weak typings in your dependencies
• more

Examples

const foo1 = returnsAny();  // error
const foo2: any = returnsAny(); // ok
const foo3: string = returnsAny(); // probably ok, programmer gave an explicit type? could also require an explicit cast.
const foo4 = returnsAny() as string;  // ok

const {foo5}: {foo5: string} = returnsAny(); // ok

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

提案とその例から始め、特に推論された any と明示的に注釈付けされた any の違いを確認してください。列挙された推論ケースを確認し、それぞれでオプションがどのように動作すべきかを定義してください。コンパイラーが、any として推論された注釈のないシンボルを、出力される JavaScript を変更せずに一貫して指摘できれば完了です。