Operator '+' cannot be applied to types 'T' and 'T'.

[Jameskmonger]

TypeScript Version: 2.1.1 / nightly (2.2.0-dev.201xxxxx)

Code

function add<T extends string | number>(x: T, y: T): T {
  return x + y;
}

add<string>("a", "b");
add<number>(5, 3);
add("a", "b");
add(5, 3);

Expected behavior:

Code to compile fine, because T is of type string | number and therefore can have the + operator applied.

Actual behavior:

Operator '+' cannot be applied to types 'T' and 'T'.

[arusakov]

Also doesn't work correctly for simple union type:

type X = number | string;

function add(a: X, b: X) {
    return a + b; // Error: Operator '+' cannot be applied to types 'X' and 'X'.
}

[DanielRosenwasser]

For those who want to tackle this, you might need to use getConstraintOfTypeParameter.

[arusakov]

@DanielRosenwasser

type U = string | number;
declare const x: U;
declare const y: U;
declare const s: string;
declare const n: number;

x + y; // resultType is U
x + s; // resultType is string
x + n; // resultType is U

Am I right?

[Jameskmonger]

@mhegazy I noticed that your comment in #13073 said this should be relaxed for x + y when x and y are both string | number, but I'm not sure that aligns correctly with the behaviour in Chrome.

function SomeClass (val) {
    this.val = val;
}

SomeClass.prototype.toString = function () {
    return `value [${this.val}]`;
};

var x = new SomeClass(3);

var z = (x + x);
console.log(z); // "value [3]value [3]"

[mhegazy]

looking at this again, i do not think this is as simple as i originally thought. we need to talk about the new rules for the operator.

[DanielRosenwasser]

type U = string | number;
declare const x: U;
declare const y: U;
declare const s: string;
declare const n: number;

x + y; // resultType is U
x + s; // resultType is string
x + n; // resultType is U

@arusakov To expand that out:

x + y; // resultType is `string | number`
x + s; // resultType is `string`
x + n; // resultType is `string | number`

that looks correct. However, if U was a type parameter whose constraint was string | number, I don't think necessarily think U would stick around.