In Javascript, if you have a function can you get the arguments by name?

Question

Let's say you have

var funct=function(a,b){
    return a+b;
};
console.log(funct);

Is there any way you can get the names of the arguments (a and b) from funct? If you are with in the function do you have any access to them? I know "arguments" gives you an array-like object, but is there anything that gives you a map-like or object-like representation of the arguments so that you can get their names when the function was declared?

You can already access the arguments to a function by name by using the actual names – it's probably best to use the argument object to implement variadic functions. If you want dynamic named arguments, use the options-hash pattern. (The one jQuery and other libraries use pervasively.) — millimoose
– millimoose, Commented Jun 14, 2012 at 16:04

gen_Eric · Accepted Answer · 2012-06-14 16:20:44Z

5

No, you cannot do this. There's no way to get the name of variable as a string.

If you really need this, I suggest instead of passing multiple parameters, pass an object.

var funct=function(args){
    var argsNames = Object.keys(args); // Get the keys of the args object
    console.log(argsNames); // ['a','b']
    return args.a + args.b;
};

Then call it like so:

funct({
    a: 12,
    b: 2
});

edited Jun 14, 2012 at 16:20

answered Jun 14, 2012 at 16:05

gen_Eric

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3 Comments

ossek Over a year ago

could make use of object.keys this way as well: developer.mozilla.org/en/JavaScript/Reference/Global_Objects/…

gen_Eric Over a year ago

@WaqarAlamgir: What are you talking about?

gen_Eric Over a year ago

@sherlocken: Thanks, I'll add that to the answer.

gen_Eric · Accepted Answer · 2012-06-14 16:22:18Z

2

One way is to turn the function into a string and parse it out

var funct=function(a,b){
    return a+b;
};
var re = /\(([^)]*)/;
var daArgs = funct.toString().match(re)[1].split(/,\s?/);
console.log(daArgs);

jsFiddle

Still have no clue why you would need it.

edited Jun 14, 2012 at 16:22

gen_Eric

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answered Jun 14, 2012 at 16:08

epascarello

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2 Comments

gen_Eric Over a year ago

I assumed he wanted the argument names inside the function, not outside.

epascarello Over a year ago

@Rocket Than replace funct.toString() with the deprecated arguments.callee.toString(), but it will not run in strict mode. :)

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