I have two function a() and b(), both are variadic functions, let say when I call function a() like this :

a(arg0, arg1, arg2, arg3, ...., argn);

then the function b() will be called as well inside a(), but without the first argument "arg0" in the arguments list of a() :

b(arg1, arg2, arg3, ...., argn);

Is there any way for it?

I have two function a() and b(), both are variadic functions, let say when I call function a() like this :

a(arg0, arg1, arg2, arg3, ...., argn);

then the function b() will be called as well inside a(), but without the first argument "arg0" in the arguments list of a() :

b(arg1, arg2, arg3, ...., argn);

Is there any way for it?

• javascript
• function
• variadic

• 1 Related: stackoverflow.com/questions/960866/… – Ray Toal Commented Jul 24, 2011 at 18:17

1 Answer 1

Every JavaScript function is really just another "object" (object in the JavaScript sense), and comes with an apply method (see Mozilla's documentation). You can thus do something like this....

b = function(some, parameter, list) { ... }

a = function(some, longer, parameter, list)
{
   // ... Do some work...

   // Convert the arguments object into an array, throwing away the first element
   var args = Array.prototype.slice.call(arguments, 1);

   // Call b with the remaining arguments and current "this"
   b.apply(this, args);
}