Object narrawing to function broken on v3.1

[dxmaxwell]

TypeScript Version: 3.2.0-dev.20181018

Search Terms: function narrowing guard

Code

// Error in 3.2.0-dev.20181018, but works in v3.0.1!
function stringify(x: {} | (() => string)): string {
    if (typeof x === 'function') {
       return x();
    }
    return String(x);
}

Expected behavior: The example code should compile without error.

Actual behavior: Compiler error: "Cannot invoke an expression whose type lacks a call signature. Type 'Function | (() => string)' has no compatible call signatures."

Playground Link: http://www.typescriptlang.org/play/#src=%2F%2F%20Error%20in%203.2.0-dev.20181018%2C%20but%20works%20in%20v3.0.1!%0D%0Afunction%20stringify(x%3A%20%7B%7D%20%7C%20(()%20%3D%3E%20string))%3A%20string%20%7B%0D%0A%20%20%20%20if%20(typeof%20x%20%3D%3D%3D%20'function')%20%7B%0D%0A%20%20%20%20%20%20%20return%20x()%3B%0D%0A%20%20%20%20%7D%0D%0A%20%20%20%20return%20String(x)%3B%0D%0A%7D%0D%0A

Related Issues: None that I could find.

[ghost]

The problem is that all functions are assignable to {}. So {} | (() => string) actually allows any function, not just () => string, so your code as-is allows stringify(() => 2);.
The following should work better:

type FunctionsMustReturnString<T> = T extends Function ? () => string : T;

function stringify<T>(x: FunctionsMustReturnString<T>): string {
    if (typeof x === 'function') {
       return x();
    }
    return String(x);
}

stringify(2); // OK
stringify(() => ""); // OK

stringify(() => 2); // Error
stringify((p: string) => p); // Error

[dxmaxwell]

Thank you for the response, I will certainly update my code as you have suggested. I guess I should consider this bug in v3.0, rather then a regression in v3.1.

[RyanCavanaugh]

Also some discussion of a similar pattern in #27278

[jack-williams]

Related? #26970

[dxmaxwell]

Something that seems inconsistent is that the following does NOT throw an error in 3.2.0-dev.20181018:

function stringify<T = {} | (() => string)>(x: T): string {
    if (typeof x === 'function') {
       return x();
    }
    return String(x);
}

Which seems equivalent to the original erroneous code?

[ghost]

When you write T = {} | (() => string) that gives a default value for T, but T doesn't actually have to be that type. It could be null which isn't even assignable to it.
For a type with no known function signatures (e.g. x: unknown), typeof x === "function" narrows x to Function which supports being called with any signature.

[dxmaxwell]

That makes perfect sense... thanks again!