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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?
258
To call a function with a variable number of arguments, simply specify any number of arguments in the function call. An example is the printf function from the C run-time library. The function call must include one argument for each type name declared in the parameter list or the list of argument types.
A variadic function is a function where the total number of parameters are unknown and can be adjusted at the time the method is called. The C programming language, along with many others, have an interesting little feature called an ellipsis argument.
It takes one fixed argument and then any number of arguments can be passed. The variadic function consists of at least one fixed variable and then an ellipsis(…) as the last parameter. This enables access to variadic function arguments. *argN* is the last fixed argument in the variadic function.
You can call a Javascript function with any number of parameters, regardless of the function's definition. Any named parameters that weren't passed will be undefined.
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);
}
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