Typescript Type Guards and fat arrow function

Question

Shouldn't this compile correctly? I get an error "Property 'hello' does not exist on type 'object'." in the highlighted line.

I can access g.hello outside the fat arrow function without problems.

class Test {
    constructor() {
    }
    hello() : string {
        return "Hello";
    }
}

let g : object;

if (g instanceof Test) {
    () => {
        g.hello();    ////// ERROR HERE /////
    };
}

Answer 1

The narrowing that a type-guard does on a variable (or anything else) will not cross fucntion boundaries. This is a design limitation.

A way to work around this issue is to assign g to a new variable, which will have it's type inferred based on the narrowing. Accessing the new variable in the arrow function will work as expected:

class Test {
    constructor() {
    }
    hello() : string {
        return "Hello";
    }
}

let g : object;

if (g instanceof Test) {
    const gTest = g;
    () => {
        gTest.hello();
    };
}

Another way to work around this issue if g does not change, is to declare g with const. This will let the compiler preserve the narrowing:

let g : object;

if (g instanceof Test) {
    const gTest = g;
    () => {
        gTest.hello();
    };
}

Playground Link