I came across something interesting today. Apparently the following code is not the same, as it gives different values of $array in the end:

PHP Code:

$array = array_merge($array, array_splice($array, $position, count($array), $element)); 


PHP Code:

$temp = array_splice($array, $position, count($array), $element);
$array = array_merge($array, $temp); 


It seems what might be happening that when a variable is passed by reference and variable is changed, the new value isn't available until the end of the statement. Why is this? Does this occur in other languages?

It seems what might be happening that when a variable is passed by reference and variable is changed, the new value isn't available until the end of the statement. Why is this? Does this occur in other languages?

Not exactly. You touched a sensible subject in PHP.
In php4, at least, every variable transfered from a function to another is in fact COPIED.
So, when you do

PHP Code:

$var="my value";
$var=str_replace(" ",'_', $var); 


It means that str_replace receive a copy of $var, works on it, and return that copy that overwrite the old $var.

Now, when you work by reference, like this:

PHP Code:

$var="my value";
$var=str_replace(" ",'_', &$var); 


str_replace don't receive a copy of $var, but the memory adresse of the value of $var.
It means that when you are working with references passed calls, you can change the content of a variable, even when using incorrect syntax.
The example up there is valid, but this would have the same effect:

PHP Code:

$var="my value";
str_replace(" ",'_', &$var); 


Because of the reference call is working on the content of the original variable.

I don't know for a lot of languages, but I've learned to use this in C (it's called a pointer, if I'm remembering right).
My guess is that it's a fairly common construction, and surely a lot of languages are providing this facility.

I know what you mean, but that's not exactly what I was referring to. I guess I should have mentioned this in my original post, but the array which is passed into array_splice() is always passed by reference.

I used a bad example because I called too many things $array.

PHP Code:

$result = array_merge($array, array_splice($array, $position, count($array), $element)); 


PHP Code:

$temp = array_splice($array, $position, count($array), $element);
$result = array_merge($array, $temp); 


The values of $result are different for both of the above examples. I've been thinking about it a little bit more, and yeah, since it is being passed by reference, it should be available immediately. I think what must be happening is that since the $array argument in the array_merge() call is being passed by reference, it's being copied for use in array_merge() before the array_splice() call is made.

Actually that makes perfect sense. I get it now.

Maybe you can try this.

PHP Code:

$result = array_merge($array, (array_splice($array, $position, count($array), $element))); 


I think you will get the same result like in second example.

Shivaji