foreach reference pit

Look at the code below

$arr1 = [1, 2];

foreach($arr1 as $key => $value) {
　　$value = $value + 1;
}

var_dump($key, $value); //结果输出1,3

Description: In the above foreach loop, when the loop is finished, the temporary variable $ key and $ value variable will not be automatically released. Value will be saved. But this time the modification does not affect the value of $ val $ arr.

Foreach understand the principle, the result of the above is easy to understand, when the foreach loop, each loop will copy the value of `$ arr1` element to the temporary variable: $ key, $ value. E.g., on the first iteration, the following

$key = 0;
$value = $arr1[0];

When the second cycle, as follows

$key = 1;
$value = $arr1[1];

 

Look forach used references

Quote: If you want to modify elements of the array as they traverse the array, you can use a reference to the foreach $ val in. At this time, $ val referenced elements of the array element points to the current memory address, that is a period of shared memory address. $ Val thus modified while changing the value $ arr [$ key] value.

of arr1 $ = [. 1, 2 ];
 the foreach ( $ of arr1  AS  $ Key => $ value ) {
     $ value = $ value +. 1 ; 
} 
var_dump ( $ Key , $ value ); // result output l, 3 

var_dump ($ of arr1); 
$ value of arr1 [ '0' => 2, '1' => 3]

Description: $ key, value and above the $ value of non-quoted case of the same. After using foreach in & references, when the end of the foreach, $ key and $ val variables will not be automatically freed, but at this time and $ val arr $ COUNT ($ arr) - 1 point to the same memory address. Therefore, at this time $ val modified value will also change the $ arr [3] value.

When foreach quote, principles and above, but is a reference copy, as the original

First cycle:
 $ Key = 0 ;
 $ value = & $ of arr1 [0 ]; 

Second cycle: 
$ Key =. 1 ;
 $ value = & $ of arr1 [. 1];

 

Look at an example:

$arr1 = [1, 2];
foreach($arr1 as $key => &$value) {
    $value = $value + 1;
}
var_dump($arr1);

$value = 100;

var_dump($arr1);

When the second output $ arr1, the value of the second element is the 100.

Understand the principles, you know pit. If we have two foreach loop, using the same temporary variable will be a problem, the use of references in a foreach time.

Such as:

of arr1 $ = [. 1, 2 ];
 the foreach ( $ of arr1  AS  $ Key => & $ value ) {
     $ value = $ value +. 1 ; 
} 
// At this time $ value = & $ arr1 [1 ], $ value changes back value will change the value of $ arr1 [1] element.
arr2 is $ = [10, 20 is, 30 ];
 the foreach ( $ arr2 is  AS  $ value ) {
     var_dump ( $ of arr1 ); 
}

 

We found that when the cycle $ arr2, $ arr2 copied to the $ element value, while the value of the last element of arr1 $ also changes, then the pointer value is equal to $ arr2 is the corresponding element at this time. Sequentially push the last value = $ arr2 elements of the last element $ arr1.

Solution, using unset () release element or the second loop array when using different temporary variables

$arr1 = [1, 2];
foreach($arr1 as $key => &$value) {
    $value = $value + 1;
}
unset($value);
$arr2 = [10, 20, 30];
foreach ($arr2 as $value) {
    var_dump($arr1);
}