Spread syntax ignores generic type constraints #46146
Description
Bug Report
🔎 Search Terms
typescript extends spread
Through this I found #26412, which may be related to this, but I can't quite tell, in part because a code example that it claims raises a compile error in TS 3.0 no longer raises a compile error in 4.4.3. There appear to be a number of bugs related to that, but as far as I can tell, none of them match the particular problem I'm having. I might have missed one, though, in which case I apologize.
🕗 Version & Regression Information
4.4.3. Also happens in 4.5.0-dev.20210930 (see playground link below).
This is the behavior in every version I tried, and I reviewed the FAQ for entries containing the word extends and spread.
⏯ Playground Link
💻 Code
type Thing = {
hello: number[];
}
function operateOnThingViaSpreadThenAssignment<T extends Thing>(thing: T): T {
const newThing = { ...thing };
// Here we are assigning a string instead of number[] to the "hello" property,
// and TS raises a compile error as expected.
newThing.hello = "hi";
return newThing;
}
function operateOnThingViaSpreadOnly<T extends Thing>(thing: T): T {
// Here we are assigning a string instead of number[] to the "hello" property,
// but TS doesn't complain.
return { ...thing, hello: "hi" };
}🙁 Actual behavior
TypeScript appears to treat the two functions differently, correctly raising a compilation error for operateOnThingViaSpreadThenAssignment() while curiously being fine with operateOnThingViaSpreadOnly().
The latter seems particularly bad because, as far as I can tell, it's effectively not reporting something that is actually a type error.
🙂 Expected behavior
I expect TypeScript to raise the same error--or at least, something very similar, along the lines of Type 'string' is not assignable to type 'number[]'--for both functions.