Spread operator function arguments (regression)

[sheley1998]

TypeScript Version: 3.2.1

Search Terms: spread, arguments

Code

function test(a: number, b: number, c: number, d: number) {
    console.log(a, b, c, d);
}

const nums = [0, 1, 2, 3];

test(...nums);

Expected behavior:
the snippet above should compile without error

Actual behavior:
Error TS2556: Expected 4 arguments, but got 0 or more

[ahejlsberg]

This is working as intended and is not a regression. We infer type number[] for nums and therefore we have to assume that the array may have as few as zero elements. It would be nice to infer a tuple type, but the reality is that arrays are mutable (the only thing that is const in the example is the array reference--the contents of the array are not). There are many similar issues already logged, e.g. #28336, #20899, #16389, #3369.

[sheley1998]

Thank you for the clarification :)

[typescript-bot]

This issue has been marked 'Working as Intended' and has seen no recent activity. It has been automatically closed for house-keeping purposes.