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

Question

Consider this TypeScript code, compiled with 2.6.1:

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

const string: string = "a";
foo(string, num => num.parseInt());

I would expect the compilation to fail, because the function foo is called with a string, but the passed callback function is using a method that is not available in string -- while the function signature indicates that type of argument in callback function should be the same as the type of the first parameter.

However, the code compiles and then fails in runtime.

What am I missing?

"is using a number method" --- there is no parseInt in the Number.prototype. — zerkms
– zerkms, Commented Nov 22, 2017 at 20:18
look at @zerkms comment. If you try foo("a", x => console.log(x.toFixed())) the compiler will tell that the property toFixed does not exist on type string — user8928802
– user8928802, Commented Nov 22, 2017 at 20:51

artem · Accepted Answer · 2017-11-22 21:12:16Z

6

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 - ...!

answered Nov 22, 2017 at 21:12

artem

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2 Comments

jcalz Over a year ago

Relevant FAQ entry

ILikeFood Over a year ago

Thanks, this fixed my issue as well!

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