如何在TypeScript中定义排除`any`的类型约束？@thisisagile/easy库Get类型优化问询

Great question! Let's tackle both parts of your query clearly:

1. Is there a TypeScript type that allows any type except any? #

TypeScript doesn't have a built-in type for this, but we can easily create a type checker to detect whether a given type is any, then use that to exclude it from our type definitions.

The key insight here is leveraging TypeScript's special handling of any: when you intersect any with any other type, the result is still any. We can use this quirk to build our IsAny utility:

// Checks if a type T is exactly `any`
type IsAny<T> = 0 extends (1 & T) ? true : false;

How this works:

• For T = any: 1 & any evaluates to any, and 0 extends any returns true (since any is compatible with all types).
• For any other type T (like string, number, unknown, etc.): 1 & T evaluates to a type that 0 does not extend, so the result is false.

2. Implementing a Get type that fixes your any edge case #

With our IsAny checker, we can modify your existing Get type to behave exactly as you described: when T is any, we only allow the function variant; otherwise, we allow both the value and function variants.

First, let's confirm the Func type (matching your library's definition):

type Func<T, Args = unknown> = (...args: Args[]) => T;

Now, the improved Get type:

type Get<T = unknown, Args = unknown> = IsAny<T> extends true
  ? Func<T, Args> // Only allow function when T is any
  : T | Func<T, Args>; // Allow both value and function otherwise

Why this fixes your issue #

In your original Get type, Get<any> resolved to any | Func<any, Args>. Since any absorbs all other types in a union, the entire type became any—this is why the compiler couldn't distinguish between a raw value and a function returning any.

With the modified type, Get<any> strictly resolves to Func<any, Args>, so the compiler will enforce that it's a function, and your Get() evaluation function will behave as expected.

Example usage #

// Valid: T is string, can be value or function
const stringValue: Get<string> = "hello world";
const stringFunc: Get<string> = () => "hello world";

// Valid: T is any, must be a function
const anyFunc: Get<any> = () => "some any value";

// Invalid: T is any, cannot be a raw value (compiler error)
const invalidAnyValue: Get<any> = "some any value";

This implementation should resolve the unexpected behavior you saw in edge cases involving any.

内容的提问来源于stack exchange，提问作者aahoogendoorn