How to pass all arguments as collection to another function and not as single argument?

Question

I have a function that accepts any number and kind of arguments, so no specific parameter has been defined. This function should call another function passing all arguments.

The problem is that I can pass "arguments" in order to include all arguments but in this case it will work like a single argument and not the way we expect arguments to work.

An example:

The main function:

function handleCall() {
   // let's call a sub-function
   // and pass all arguments (my question is how this is handled the right way)
   function callSubFunction( arguments );
}

function callSubfunction( userid, customerid, param) {
   // passed arguments are now 
   alert( 'userid = ' + userid );

   // this will not work, you have to use arguments[2]
   alert( param );
  }

The example call:

handleCall( 1029, 232, 'param01' );

Using the approach above, all arguments will be stored in "userid" as pseudo-array and items can be accessed e.g. arguments[2] but not using the parameter name "param".

In ColdFusion, the solution for such stuff is the parameter "argumentCollection", this way you can pass parameters stored in a structure without being converted to a single argument with the type struct containing all key/values.

How can I achieve the same with JavaScript?

score 78 · Answer 1 · answered Jan 25 '17 at 09:51

78

If you want to do the same with the spread syntax, you can use the following:

function handleCall(...args) {
    callSubFunction(...args);
}

answered Jan 25 '17 at 09:51

alexmngn

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For typescript, we can extract the type of our function parameters and use that to type the rest args: type handleCallParams = Parameters; – chrismarx May 14 '21 at 12:27
That's the good stuff! – Charney Kaye Aug 08 '21 at 07:47
For typescript with a constructor and a call to super, you can use `ConstructorParameters` – Jespertheend Sep 14 '21 at 19:36

score 48 · Accepted Answer · edited Jan 08 '18 at 19:20

48

You can use the .apply() method to call a function and pass the arguments as a set.

callSubFunction.apply( this, arguments );

The first argument will set the value of this in the allSubFunction method. I just set it to the current this value. The second is the collection of arguments to send.

So your handleCall() function will look like:

function handleCall() {
     //set the value of "this" and pass on the arguments object
    callSubFunction.apply( this, arguments );
}

It isn't required that you send an Arguments object. You could send an Array of arguments if the circumstance required.

edited Jan 08 '18 at 19:20

Johann

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answered Jan 16 '11 at 15:44

user113716

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how to do it in ES6 function? – IsmailS Jan 17 '20 at 14:45
1

The link to the documentation here is **dead**. [My edit](https://stackoverflow.com/suggested-edits/5041911), which correctly **fixed** the links, was rejected by two users with *the same* copy-pasted message about "readability", despite the message not affecting visible text. (I suspect they were farming/botting for site rep.) In the meantime, here is a working link to MDN's `Function.prototype.apply()`: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/apply – JamesTheAwesomeDude Jan 27 '21 at 22:57

score 1 · Answer 3 · answered Jan 16 '11 at 15:46

1

Use apply like so:

function Foo()
{
    Bar.apply(this, arguments);
}

function Bar(a, b)
{
    alert(a);
    alert(b);
}

answered Jan 16 '11 at 15:46

Jimmy Chandra

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How to pass all arguments as collection to another function and not as single argument?

3 Answers3

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