Intersection with type parameter doesn't subtract properties

[smoogly]

TypeScript Version: 2.6.1 && 2.7.0-dev.20171207

Code

interface InternalProps {
    a: number;
}
interface ExternalProps {
    b: string;
}

function make<T = {}>(arg: T & InternalProps): T & ExternalProps {
    return 1 as any;
}

const x = make<{}>({a: 1});
console.log(x.a); // Correct: type error

const y = make({a: 1});
console.log(y.a); // Incorrect: no type error

Expected behavior:
Property a should not be expected to exist on y.
Both examples should fail.

Actual behavior:
Only first example is considered a type error.

[ghost]

Looks like a duplicate of #20505.

[smoogly]

I'm not sure, but I don't think it's a duplicate.
In this case looks like T is inferred to be InternalProps, which is unexpected:

interface InternalProps {
    a: number;
}
interface ExternalProps {
    b: string;
}

function make<T = {}>(arg: T & InternalProps): T & ExternalProps {
    return 1 as any as T & ExternalProps;
}

const x = make<{}>({a: 1});
console.log(x.a); // Correct: type error
console.log(x.blah); // Correct: type error

const y = make({a: 1});
console.log(y.a); // Incorrect: no type error
console.log(y.blah); // Correct: type error

./node_modules/.bin/tsc test.ts                                                                            
test.ts(13,15): error TS2339: Property 'a' does not exist on type 'ExternalProps'.
test.ts(14,15): error TS2339: Property 'blah' does not exist on type 'ExternalProps'.
test.ts(18,15): error TS2339: Property 'blah' does not exist on type '{ a: number; } & ExternalProps'.

[ghost]

The thing they have in common is that adding a default to a function's type parameter doesn't turn off type inference. So if a type can be inferred, that type will be used instead of the default.

But in this case I see that T could be inferred as {} since the input is T & InternalProps and InternalProps can handle the a property. One alternative to subtraction types (#4183) that I've heard discussed in the past I think looked something like this:

declare function removeA<T>(arg: T & { a: any }): T;
const o = removeA({a: 1});
o.a; // Should be an error?

[smoogly]

@andy-ms, yes if the example you show was implemented,
my scenario would produce the results I expect as well.

[mhegazy]

The compiler can not make these assumptions that the property exclusively exists in one constituent of the intersection.. i.e. T & { a: number } does not guarantee that T itself does not have a property "a" with the same type or a different one. I think what you need is Diff (#19569).

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.