So with this code:
$x = 'hello world';
$y = 'bye world';
$arr = array ($x, $y);
$arr[0] = 'new string';
var_dump($x);
$arr[0] = 'new string'; it's not changing the value of $x. How can I change $x value referencing the array position?
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print_r($arr) you are going to replace array[0] index valueMD. Jubair Mizan– MD. Jubair Mizan2020-07-30 11:02:54 +00:00Commented Jul 30, 2020 at 11:02
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2 Answers
The value of $x will be copied to the array! There is no reference from $arr[0] to $x!
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You could create a reference to $x, instead of copying the variable. When passing it into the array as you are, you are merely copying the value of $x, and inserting that copy into the array. If you instead pass $x as a reference, using &$x, any updates made will be made to $x instead. Since the value is a reference, it will be updated "both places".
$x = 'hello world';
$y = 'bye world';
$arr = array (&$x, $y);
$arr[0] = 'new string';
var_dump($x);
Output:
string(10) "new string"
- Live demo at https://3v4l.org/KSeKg
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