打字稿通用类型断言

Question

So Here is a summary of my observation working with typescript.

Here is some code:

type someTypeEnum = '1';
type someOtherTypeEnum = '2' | '3';
type combinedTypeEnum = someTypeEnum | someOtherTypeEnum;

Here is the first case :-

function typeAssertion<T extends combinedTypeEnum>(args: T): args is someTypeEnum {
    // The error i get
    // A type predicate's type must be assignable to its parameter's type.
    //  Type '"1"' is not assignable to type 'T'.
    return undefined;
}

I cannot understand why this thing fails coz we have already limited our arguments to combinedTypeEnum, in case we do

typeAssertion('4')

We already get an error stating '4' is not a valid argument so why is it that args is someTypeEnum is considered an in-valid predicate.

Here is the second case :-

function typeAssertion(args: combinedTypeEnum): args is someTypeEnum {
    return undefined;
}

This seems to work fine but in case we do this :-

function someFunction<T extends combinedTypeEnum>(args: T): T {
    if (typeAssertion(args)) {
        // args here is  'T & "1"' 
        args
    }
    return args
};

why is it that we have T & "1" and not only "1", we specifically asserted that it is someTypeEnum.

I was really curious as to why such decisions were made. It would be really helpful to see how things break in case things were done in a different manner.

Answer 1

extends doesn't make much sense when you have string literals. To make the explanation easier let me use other types. Consider these three classes:

class Animal {}

class Dog extends Animal {}

class Cat extends Animal {}

when we use generics the actual type is set by the caller:

function foo<T extends Animal>(arg: T) {}

foo(new Dog()); //T is Dog, equivalent to foo(arg: Dog) {}
foo(new Cat()); //T is Cat, equivalent to foo(arg: Cat) {}

Now you may already see where we are going. Let's use a type predicate:

function foo<T extends Animal>(arg: T): arg is Cat {}

When we call foo(new Dog()) the last example becomes equivalent to this:

function foo(arg: Dog): arg is Cat {}

And of course it doesn't work or make sense.

As for your second example: The type of the variable doesn't change. The point is that by asserting a specific type the compiler allows you to do whatever can be done with this type.

Answer 2

UPDATE:

Or much simpler:

function typeAssertion(args: combinedTypeEnum): args is someTypeEnum {
    return args === "1";
}

See this playground

---

Original:

This had me stumped for a long while as well. Actually the solution (at least in 2021, not sure if it was back when the question was asked) is:

function typeAssertion<T extends combinedTypeEnum>(args: T): args is T & someTypeEnum {
    return args === "1";
}

The idea behind this (as far as I understand it from this answer ) is this: when you call typeAssertion("2") , the T gets the value "2" (literal type "2" ), meaning that you end up with the function:

function typeAssertion(args: "2"): args is someTypeEnum

which obviously doesn't make sense. I'm not sure the workaround (with T & ) makes more sense, but it works:

type someTypeEnum = '1';
type someOtherTypeEnum = '2' | '3';
type combinedTypeEnum = someTypeEnum | someOtherTypeEnum;

function typeAssertion<T extends combinedTypeEnum>(args: T): args is T & someTypeEnum {
    return args === "1";
}

const a: combinedTypeEnum = "1"
const b: combinedTypeEnum = "2"
const c: combinedTypeEnum = "3"
const d = "1"
const e = "2"
const f = "4"

let one: "1" = "1"

if (typeAssertion(a)) one = a
if (typeAssertion(b)) one = b
if (typeAssertion(c)) one = c
if (typeAssertion(d)) one = d
if (typeAssertion(e)) one = e
if (typeAssertion(f)) one = f // this one gives an error

See in Playground