TypeScript: types for variadic functions [duplicate]

Question

Possible Duplicate:
open-ended function arguments with TypeScript

Is there any acceptable type signature for variadic functions? Example:

function sum () {
  var sum = 0;
  for (var i = 0; i < arguments.length; i++) {
    sum += arguments[i];
  }
  return sum;
};

console.log(sum(1, 2, 3, 4, 5));

gives me compilation error:

foo.ts(9,12): Supplied parameters do not match any signature of call target

See stackoverflow.com/questions/12697275/… - open-ended function arguments with TypeScript — Ryan Cavanaugh
– Ryan Cavanaugh, Commented Oct 2, 2012 at 22:17

peterh · Accepted Answer · 2017-02-17 10:25:40Z

70

In TypeScript you can use "..." to achive the above pattern:

function sum (...numbers: number[]) {
  var sum = 0;
  for (var i = 0; i <  numbers.length; i++) {
    sum += numbers[i];
  }
  return sum;
};

This should take care of your error.

edited Feb 17, 2017 at 10:25

peterh

1

answered Oct 2, 2012 at 22:28

mohamed hegazy

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4 Comments

Lee Whitney III Over a year ago

I guess it doesn't support the multiple type feature? For example, this will not compile: foo(...item: string[] | number[])

Borsunho Over a year ago

@Lee it won't swallow string[] | number[], but (number|string)[] is perfectly ok (although you will likely need to do some casting inside function body).

lealceldeiro Over a year ago

or ...numbers: Array<number | string>

River Tam Over a year ago

you could also make it a template function if you want to specify they're all the same: function sum<T extends number | string>(...item: Array<T>): T

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