Look at the following code:
var x = [1, 2, 3], y;
y = x;
y.splice(0,0,4);
which gives:
y = [4, 1, 2, 3] // (correct)
x = [4, 1, 2, 3] // (why did this change too?)
Why did the x array change when I called .splice() on y?
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What language is this?Paul Boddington– Paul Boddington2016-04-10 01:48:55 +00:00Commented Apr 10, 2016 at 1:48
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Answered pretty well by stackoverflow.com/questions/7480654/…Dawson Toth– Dawson Toth2016-04-10 02:07:07 +00:00Commented Apr 10, 2016 at 2:07
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ok, so how do you keep x the same ?pomeKRANK– pomeKRANK2016-04-10 02:09:50 +00:00Commented Apr 10, 2016 at 2:09
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1 Answer
Objects (including arrays) are passed by reference (effectively, this is what it does... -- purists may disagree with this statement). The splice method mutates the original array. Therefore, since x and y point at the same array, a splice on y will change x as well. To do a shallow clone of x in to y, do y = x.slice(). (Note that any objects within x won't be cloned; they'll be passed by reference.)
var a = [1,2,3];
var b = a;
a[0] = 42;
alert(b[0]); // will show 42
var c = a.slice(); // explicitly makes a copy
a[1] = 6502;
alert(c[1]); // will show 2, not 6502
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