Php foreach- Why it doesn't override the array value while iterating?

PHP uses a copy on write or lazy copy mechanism with reference counting for memory handling.

In the first stage of foreach iteration, $arr is "soft copied" (no actual copy, but only the refcount of the zval of $array is increased). checkout variable container and label of PHP variables.

In our case,

• it first keep a reference to actual $arr variable, and the first value of array is fetched from the reference point.
• For later use first array element is assigned to second $arr foreach($arr as $arr)
• In the next iteration, second array value is fetched from previously saved reference point, and that value is assigned to same $arr and continues.
• As value of $arr is changed inside loop, you cannot print that array outside loop using $arr variable.

Hope you can understand...

For an in-depth study checkout http://www.php.net/manual/en/internals2.opcodes.fe-fetch.php and http://www.php.net/manual/en/internals2.opcodes.fe-reset.php

Interestingly you can play with this too...

current() inside a foreach loop will always returns the same element!!!

Code

$source = array(10, 20, 30);
foreach ($source as $value) {
    echo 'Value is ', $value, ', and current() is ', current($source), '<br>';
}

Output

Value is 10, and current() is 20
Value is 20, and current() is 20
Value is 30, and current() is 20

foreach ($arr as $arr) {
    // code
}

This may be confusing to you because your mental model of how this works is:

1. PHP takes the current element out of $arr and advances the array pointer
2. it assigns the value to $arr
3. it executes // code
4. it repeats from step one

What it actually does though is this (or at least an approximate simplification of it):

1. PHP starts the foreach operation, taking the array value that is associated with $arr
2. it puts this array somewhere internally, let's say a var iterator
3. it takes the current element from var iterator and advances the iterator
4. it assigns the value it took to $arr in your script
5. it executes // code
6. it repeats from step 3

In other words, it does not take the next value from $arr on every iteration, it only used this once to get the initial iterator. The actual thing that it iterates over is stored internally and is not associated with $arr anymore. That means $arr is free for other purposes in your script and PHP is happy to reuse it on each iteration. After the loop is done, the last element remains in $arr.

If you know anonymous functions, maybe this illustrates better why this works the way it works:

foreach($arr, function ($arr) {
    // code
});

The implementation of such a foreach function can be simply this:

function foreach(array $array, callable $code) {
    while (list(, $value) = each($array))  {
        $code($value);
    }
}

$arr and $arr are same in your code.Source array and reference variable are having same name.So foreach keeps the value that was last recieved it doesnt relinquishes the variable.

foreach use a copy of the original array, so you could think the code

foreach($arr as $arr){
    echo $arr.'<br>';
}

is just like below:

for ($i = 0, $arr_tmp = $arr, $len = count($arr_tmp); $i < $len; $i++) {
  $arr = $arr_tmp[$i];
  echo $arr.'<br>';
}

This explains why the $arr is 3 at last.

you are over writing the array when you fetch a record , you are asking the code to fetch a record from $arr and put it into the array named $arr so you are over writing the whole array with single parameter before $arr = {1,2,3}; after this line and first iteration foreach ($arr as $arr) now the the $arr is equal to 3 which the last element of the array , i hope you under stand the problem

Hmm, after a bit of investigating I came up with this code:

<?php
$arr=array(1,2,3);

echo '$arr value : '. $arr;

echo '<br>';

foreach($arr as $arr){
    echo $arr.'<br>';
}

foreach($arr as $key) {
    echo $key.'<br>';
}
echo '$arr value : '. $arr;

Which had this output:

$arr value : Array
1
2
3
PHP Warning:  Invalid argument supplied for foreach() in ...file...

My read on this is that the second $arr is shadowing the first one while inside the foreach loop, but as soon as it exits the foreach loop it internally unsets the symbol ($arr)

foreach first evaluates the left side before iterating over its elements. Therefore, the reassignment to the name $arr does not modify $arr.

Furthermore, the loop variable leaks into the outer context, as documented:

Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().

Of course, to avoid the confusion, you should simply use a different name for the loop variable than for the array you're looping over:

$arr=array(1,2,3);
foreach($arr as $a) { // Note the "as $a", not "as $arr"
    echo $a.'<br>';
}
echo '$arr value : '. $arr;