Narrowing from "any" using instanceOf will overwrite another narrowing expression in "||"

[yuit]

TypeScript Version: build from master 8/17

function foo(x: any) {
    if (typeof (x) === 'number' || x instanceof Number) {
        isNaN(x)  // error x has type Number
    }
}

Note: this doesn't happen if parameter x has type number | Number

[RyanCavanaugh]

But this is still an error if we correctly narrow to number | Number -- isNaN(new Number(3)) is not permitted (though maybe it should? Hard to say)

[yuit]

@RyanCavanaugh true; Though the error message is confusing then since you will be like why the type is only number

[weswigham]

isNaN coerces to a number with valueOf before checking (isNaN({}) === true), so it's conceivable that it should just have a wider argument type, rather than strictly number.

[DanielRosenwasser]

@weswigham see #4002 and #3947.

[ahejlsberg]

Since number is a subtype of Number, there's really no difference between number | Number and Number. In control flow analysis we perform subtype reduction at junction points only when the control flow type includes "foreign" types, i.e. types that don't occur in the declared type. The type guards in the if statement in combination produce the type number | Number, which then gets subtype reduced to Number if the body of the if statement. But this subtype reduction doesn't happen when the declared type itself is number | Number because the types aren't considered foreign.