error TS2365: Operator '+' cannot be applied to types 'string | number' and 'string | number'.

[borisovg]

Bug Report

Code does not compile.

🔎 Search Terms

"Operator '+' cannot be applied to types".

🕗 Version & Regression Information

4.7.4

💻 Code

const foo = { foo: 123, bar: 'bar' };

for (const k of ['foo', 'bar'] as const) {
  const val = foo[k];
  console.log(val + val);
}

🙁 Actual behavior

src/test.ts:5:15 - error TS2365: Operator '+' cannot be applied to types 'string | number' and 'string | number'.

🙂 Expected behavior

No error.

[RyanCavanaugh]

Operations are evaluated according to the types of their operands, not their identities. The left-hand type is string | number, the right-hand type is string | number, and the rules for + disallow an ambiguous operation like that. Whether or not the LHS and RHS refer to the same variable (or what "same" would even mean apart from bare identifiers) isn't an input to the algorithm, nor is it clear how that could be generalized.

[kelvinsekx]

Experienced something similar.

function universalAdd(monoid1: string|number){
  return monoid1 + monoid1;
}

It took me sometime to understand your statement "types of their operands, not their identities". But I got it thanks @RyanCavanaugh

[kelvinsekx]

can we close this issue already?

[haptn]

Operations are evaluated according to the types of their operands, not their identities. The left-hand type is string | number, the right-hand type is string | number, and the rules for + disallow an ambiguous operation like that. Whether or not the LHS and RHS refer to the same variable (or what "same" would even mean apart from bare identifiers) isn't an input to the algorithm, nor is it clear how that could be generalized.

I'm sorry but I'm still not clear on the problem here. If I replace the union type of string | number to type any, things will work normally.
So what I'm confusing here is that does the + operation only accepts inputs with single types (number, string, any), but not union type (string | number)?

Thank you!

[nishat-md-r-rafi]

If you are sure about the type by type guard operator "typeof". and then assume the type using "as"