Is JavaScript support default parameters in its functions?

Question

I want to make default parameter in javascript , so can i ?

jm.toInt = function (num, base=10) {
    return parseInt(num,base);
}

https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Functions/default_parameters ... see the browser support... — Arun P Johny, Sep 14 '15 at 12:27

score 4 · Answer 1 · answered Sep 14 '15 at 12:33

4

With a logical or, default values are possible.

jm.toInt = function (num, base) {
    return parseInt(num, base || 10);
}

answered Sep 14 '15 at 12:33

Nina Scholz

score 3 · Answer 2 · answered Sep 14 '15 at 12:32

ofcourse there is a way!

function myFunc(x,y)
{
   x = typeof x !== 'undefined' ? x : 1;
   y = typeof y !== 'undefined' ? y : 'default value of y';
   ...
}

in your case

    jm.toInt = function(num, base){
       return parseInt(num, arguments.length > 1 ? base: 'default value' );
    }

score 2 · Answer 3 · answered Sep 14 '15 at 12:28

2

ES6 support default parameters, but ES5 not, you can use transpilers (like babel) to use ES6 today

answered Sep 14 '15 at 12:28

Oleksandr T.

score 2 · Answer 4 · answered Sep 14 '15 at 12:29

2

It is part of ES6, but as of now, not widely supported so you can do something like

jm.toInt = function(num, base) {
  return parseInt(num, arguments.length > 1 ? base : 10);
}

answered Sep 14 '15 at 12:29

Arun P Johny

Is there any advantage to using `arguments.length > 1 ?` rather then simply `base ?` – Alexandru Severin Sep 14 '15 at 12:31
@AlexandruSeverin none... you can use any of the above formats in this case.... – Arun P Johny Sep 14 '15 at 12:36

score 0 · Answer 5 · answered Sep 14 '15 at 13:01

0

Use typeof to validate that arguments exist (brackets added to make it easier to read):

jm.toInt = function (num, base) {
    var _base = (typeof base === 'undefined') ? 10 : base
    return parseInt(num, _base);
}

answered Sep 14 '15 at 13:01

slomek

5 Answers5