javascript - Array.prototype.slice.call(arguments) vs. Array.prototype.slice.apply(arguments) - Stack Overflow

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Previous posts have talked about how Array.prototype.slice.call(arguments) work but I don't get why you're using call instead of apply when apply is used for array-like objects whereas call is used on lists of objects separated by mas. Isn't arguments an array-like object that should used apply instead of call?

Previous posts have talked about how Array.prototype.slice.call(arguments) work but I don't get why you're using call instead of apply when apply is used for array-like objects whereas call is used on lists of objects separated by mas. Isn't arguments an array-like object that should used apply instead of call?

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Share Improve this question Follow edited Sep 28, 2016 at 0:21 stackjlei asked Sep 28, 2016 at 0:05 stackjleistackjlei 10k1919 gold badges7373 silver badges124124 bronze badges 6

• 3 Even if you used apply instead of call, arguments is the thisArg, not the argsArray, so it wouldn't change anything. – user2357112 Commented Sep 28, 2016 at 0:10
• 1 Were you intending to link to this post? how does Array.prototype.slice.call() work? – Jonathan Lonowski Commented Sep 28, 2016 at 0:19
• 2 The 1st argument passed to each method is the same. The difference between them is in the 2nd or further arguments. .call(thisArg, arg0, arg1, ...) vs. apply(thisArg, argsList). The value given for thisArg bees the value of this used by the function being invoked (slice). – Jonathan Lonowski Commented Sep 28, 2016 at 0:21
• 2 Side note: Array.from() was added to a recent edition of the language to in part replace this use of slice. – Jonathan Lonowski Commented Sep 28, 2016 at 0:27
• 1 Possible duplicate of What is the difference between call and apply? – Felix Kling Commented Sep 28, 2016 at 0:28

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2 Answers 2

Sorted by: Reset to default 4

If you'd like to pass arguments to slice in array instead of one by one, then there is a difference. You could do it this way

[1, 2, 3, 4, 5, 6, 7]  ---- our example arguments
Array.prototype.slice.call(arguments, 2, 5);

here you pass to slice method 2 and 5 and these are slice arguments to indicate that you want to get items at index 1 to item at index. 4. So most likely it will be 3, 4, 5.

Array.prototype.slice.apply(arguments, [2, 5]);

This does the same but arguments for slice can be passed in an array.

If x is an array, then these are the same:

// find the "slice" method of an array and call it with "x" as its "this" argument
x.slice();

// find the "slice" method of an array and call it with "x" as its "this" argument
Array.prototype.slice.call(x);

The first argument of func.call is this this argument, or, the value of this inside the function. The remaining arguments are the arguments of the function (in this case there are none).

The slice method, called with no arguments, simply creates a new array with the same contents. That's what we want to do when we do Array.prototype.slice.call(arguments), however, since arguments isn't an array and therefore has no slice method attach to itself, we have to use the second of the two methods to get the slice method of an array, and then apply that to something that's not an array:

// find the "slice" method of an array and call it with "arguments",
// which is *not* an array, as its "this" argument
Array.prototype.slice.call(arguments);

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