Why doesn't TypeScript enforce a generic parameter in callback?

Question

Consider this TypeScript code, compiled with 2.6.1:

function foo (bar: T, baz: (T) => void) {
    const test: T = bar;
    baz(test);
}

const st         


        

Answer 1

Well, because T in baz: (T) => void is not a type name, it's a parameter name.

When you fix the syntax to mean what you want it to mean, you get the expected error:

function foo<T> (bar: T, baz: (t: T) => void) {
    const test: T = bar;
    baz(test);
}

const s: string = "a";
foo(s, num => num.parseInt()); 
// Property 'parseInt' does not exist on type 'string'.

Granted, it's really hard to spot errors like this one - I saw it only when I pasted your code into typescript playground and turned on --noImplicitAny. (T) immediately got highlighted with Parameter 'T' implicitly has an 'any' type. Even that error was puzzling for a moment - wait what - T is not a parameter, it's a type - ...!