5

In the following JS code, why doesn't f3(arr2) change the value of arr2 like f2(arr1) did to arr1? Is there any way to make f3 work as expected (if possible, without returning the modified array)?

var arr1 = [1, 2, 3, 4];
var arr2 = [1, 2, 3, 4];

function f1() {
    return [2, 3, 4, 5];
}

function f2(arr) {
    arr.push(5);
}

function f3(arr) {
    arr = f1();
}

f2(arr1);
console.log(arr1); // [ 1, 2, 3, 4, 5 ]

f3(arr2);
console.log(arr2); // [ 1, 2, 3, 4 ], expect [2, 3, 4, 5]
xngtng
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  • arr in f3 is just a reference, you should `splice` or use similar `Array.prototype` method to change the array it references. – blld May 16 '16 at 09:31

4 Answers4

3

If you want to modify the array, then you actually have to modify the array. You can't just write a reference to a different array over the variable (as that just throws away the local reference to that array).

function f3(arr) {
    arr.length = 0; // Empty the array
    f1().forEach(function (currentValue) { arr.push(currentValue); });
}
Quentin
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2

quote: "console.log(arr2); // [ 1, 2, 3, 4 ], expect [2, 3, 4, 5]"

The reason that you are not getting what you are expecting is this part here

function f3(*arr*) { *arr* = f1(); }

You are assigning the array [2,3,4,5] to the argument-name arr of the function f3, not to the arr2. Arr2 remains of course untouched and in its original state throughout your script.

function f3(*arr*) { *arr2* = f1(); } will do it.

But this answer is not my final. This is only how it appears.

Bekim Bacaj
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1

You could do it in a single step:

Array.prototype.splice.apply(arr, [0, arr.length].concat(f1()));


var arr1 = [1, 2, 3, 4];
var arr2 = [1, 2, 3, 4];

function f1() {
    return [2, 3, 4, 5];
}

function f2(arr) {
    arr.push(5);
}

function f3(arr) {
    Array.prototype.splice.apply(arr, [0, arr.length].concat(f1()));
}

f2(arr1);
document.write('<pre>' + JSON.stringify(arr1, 0, 4) + '</pre>');

f3(arr2);
document.write('<pre>' + JSON.stringify(arr2, 0, 4) + '</pre>');
Nina Scholz
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  • this is not what op was asking – Bekim Bacaj May 16 '16 at 10:10
  • @BekimBacaj, what is op expecting/asking? – Nina Scholz May 16 '16 at 10:57
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    as it seems, he is expecting a JavaScript object being assigned to a (local) variable name of the function argument (arr), to be reflected in the pointing direction that his global arr2 is pointing at. And he's asking why? The answer is: your local pointer has no way of changing the pointing direction of the one it learnt it from. Because arr2 is simply telling to the local arr where is the [1,2,3,4] object. You can't change the source of information by changing the content of the carrier. – Bekim Bacaj May 16 '16 at 22:51
-1

When you pass anything (Whether that be an object or a primitive), all javascript does is assign a new variable while inside the function... just like using the equal sign (=)

How that parameter behaves inside the function is exactly the same as it would behave if you just assigned a new variable using the equal sign.. Take these simple examples. You can ref: Link

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Quốc Hồ Thanh
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