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.

从提案及其示例开始，尤其关注推断出的 any 与显式注释的 any 之间的区别。检查列出的推断情况，并定义该选项在每种情况下应如何运行；完成的标准是编译器能够一致地标记未注释但被推断为 any 的符号，同时不改变生成的 JavaScript。