function foo(x:number, y:number, z:number) {
console.log(x,y,z);
}
var args:number[] = [0, 1, 2];
foo(...args);
Why am i getting getting this error in Typescript Playground???
Supplied parameters donot match any signature of call target.
3 Answers
So there is a little clause you may have missed:
Type checking requires spread elements to match up with a rest parameter.
Without Rest Parameter
But you can use a type assertion to go dynamic... and it will convert back to ES5 / ES3 for you:
function foo(x:number, y:number, z:number) {
console.log(x,y,z);
}
var args:number[] = [0, 1, 2];
(<any>foo)(...args);
This results in the same apply function call that you'd expect:
function foo(x, y, z) {
console.log(x, y, z);
}
var args = [0, 1, 2];
foo.apply(void 0, args);
With Rest Parameter
The alternative is that it all works just as you expect if the function accepts a rest parameter.
function foo(...x: number[]) {
console.log(JSON.stringify(x));
}
var args:number[] = [0, 1, 2];
foo(...args);
I think @Fenton explains it very well but I would like to add some more documentation and possible solutions.
Solutions:
Function overload. I prefer this solution in this case because it keeps some kind of type safety and avoids ignore and any. The original method and function call does not need to be rewritten at all.
function foo(...args: number[]): void
function foo(x: number, y: number, z: number) {
console.log(x, y, z);
}
var args: number[] = [0, 1, 2];
foo(...args);
Use @ts-ignore to ignore specific line, TypeScript 2.3
function foo(x: number, y: number, z: number) {
console.log(x, y, z);
}
var args: number[] = [0, 1, 2];
// @ts-ignore
foo(...args);
Use as any.
function foo(x: number, y: number, z: number) {
console.log(x, y, z);
}
var args: number[] = [0, 1, 2];
(foo as any)(...args);
Link with documentation regarding the spread operator:
Discussions regarding this:
function foo(x:number, y:number, z:number) {
console.log(x,y,z);
}
var args:[number, number,number] = [0, 1, 2];
foo(...args);
You can try this. Personally, I think it is the answer that best fits your question scenario.
And Mr.@anask provides a better method, more clear, easier and easier to maintain. This should also be the best practice for TypeScript features.
Use as const to automatically infer static types
Demo in TSPlayground by Mr.@anask
function foo(x: number, y: number, z: number) {
console.log(x, y, z);
}
// here young !
var args = [0, 1, 2] as const;
foo(...args);
