All you know that arguments is a special object that holds all the arguments passed to the function.
And as long as it is not an array – you cannot use something like arguments.slice(1).
So the question – how to slice everything but first element from arguments?
UPD:
seems like there is no way without converting it to an array with
var args = Array.prototype.slice.call(arguments);
If someone posts another solution it would be great, if not – I’ll check the first one with the line above as an answer.
Meddling with array functions is not actually necessary.
Using rest parameter syntax ...rest is cleaner and more convenient.
Example
function argumentTest(first, ...rest) {
console.log("First arg:" + first);
// loop through the rest of the parameters
for (let arg of rest) {
console.log("- " + arg);
}
}
// call your function with any number of arguments
argumentTest("first arg", "#2", "more arguments", "this is not an argument but a contradiction");…Rest
- See the example Fiddle
- See MDN Documentation page
###
Q. How to slice everything but first element from arguments?
The following will return an array containing all arguments except the first:
var slicedArgs = Array.prototype.slice.call(arguments, 1);
You don’t have to convert arguments to an array first, do it all in one step.
###
You can “slice without slicing” by procedurally walking the arguments object:
function fun() {
var args = [];
for (var i = 1; i < arguments.length; i++) {
args.push(arguments[i]);
}
return args;
}
fun(1, 2, 3, 4, 5); //=> [2, 3, 4, 5]
###
From https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments:
You should not slice on arguments because it prevents optimizations in
JavaScript engines (V8 for example). Instead, try constructing a new
array by iterating through the arguments object.
So Paul Rosiana’s answer above is correct
###
This can be a way:
var args = Array.from(arguments).slice(1);
###
You can use the method [].slice.call(arguments, 1)
[].slice will return you the slice function object and you can call it as the arguments and 1 are the parameters
###
You can use …rest within the function to separate the first and the rest of the arguments:
function foo(arr) {
const [first, ...rest] = arguments;
console.log(`first = ${first}`);
console.log(`rest = ${rest}`);
}
//Then calling the function with 3 arguments:
foo(1,2,3)