CODEWITHSUNDEEP
phpreference
Stack Overflow
Stack Overflow

Reputation: 2968

References are not pointer in php?

I am new to php and learning it from php.net. Can anyone tell in the summary why does that page (http://php.net/manual/en/language.references.arent.php) says What references are not & References are not pointers.

I am beginner so please can anyone explain me in simple and easy words ?

Upvotes: 2

Views: 1337

Answers (3)

Alireza Rahmani Khalili
Alireza Rahmani Khalili

Reputation: 2954

I think the code below can tell the difference between PHP reference and C pointer:

In PHP:

<?php
    $a = 0;
    $b = &a;
    echo $a; //0
    unset($b); // unset $b
    echo $a; //0 it's OK
?>

In C:

#include <stdio.h>
int main(int argc, char const *argv[]) {
    int a = 0;
    int* b = &a;

    printf("%i\n", a); //0
    free(b);              // free b
    printf("%i\n", a); //get error: *** error for object 0x7fff6350da08: pointer being freed was not allocated
}

Upvotes: 1

axiac
axiac

Reputation: 72226

In simple words, references are aliases.

A variable in PHP is stored in two pieces: the name and the value. The name points to the value.

$x = 2;
$y = $x;
$z = &$x;

When $x = 2; is executed, the name x is stored in the symbol table of the current scope and the value 2 is stored in a zval (don't ask, it's the internal name of a data structure that stores a value in PHP).

When $y = $x; is executed, the name y is stored in the symbol table of the current scope and the value of $x (2) is copied into a new zval structure.

When $z = &$x; is executed, the name z is stored in the symbol table of the current scope but a new zval is not created. Instead, z is set to point to the same zval as x.

The memory used by the variables $x, $y and $z looks like this:

+---------+                +---------+
|    x    | -------------> |    2    |
+---------+                +---------+
                                ^
+---------+                     |      +---------+
|    y    | -------------------------> |    2    |
+---------+                     |      +---------+
                                |
+---------+                     |
|    z    | --------------------+
+---------+

When a value is passed by reference to a function or a function returns a reference the same things happen, only the names are stored in different symbol tables (remark the "current scope" in the explanation above).

Let' see this code:

function f(& $z) {
    $y = $z;
    $z = $z + 2;
}
$x = 2;
f($x);

After $x = 2; the memory looks like this:

+---------+                +---------+
|    x    | -------------> |    2    |
+---------+                +---------+

During the execution of function f(), the memory looks like this:

+===== global ====+
|   +---------+   |            +---------+
|   |    x    | -------------> |    4    |
|   +---------+   |            +---------+
+=================+                 ^
                                    |
+====== f() ======+                 |
|   +---------+   |                 |      +---------+
|   |    y    | -------------------------> |    2    |
|   +---------+   |                 |      +---------+
|                 |                 |
|   +---------+   |                 |
|   |    z    | --------------------+
|   +---------+   |
+=================+

y and z are stored in a different symbol table than x and they are removed (together with the entire symbol table that contains them) when the call to f() returns.

When y is removed, its value is also removed because there is no name that points to it any more. But, because the value pointed by z is also pointed by x ($z is an alias), the value is not removed together with z and it survives the function call. f() modifies the value using $z; and this change is visible in the main program through the variable $x.

The things happen in a similar way when a function returns a reference. The function returns a value that is not copied but a new name that points to it is created into the symbols table of the calling code.

Upvotes: 6

kiks73
kiks73

Reputation: 3758

I suggest you to study Language C pointers if you have any doubt about what they are.

Anyway, this is a very simplified explanation:

References are not the memory addresses that point variables, but they are only a way to pass a parameter to a function and return that parameter modified if the function alter it.

Upvotes: 0

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