php arrays are not reference types, but value types; different variables (copies) will be created when the array is assigned, so modifying another variable will not affect the array itself. And PHP array transfer is value transfer; when calling a function, assigning the PHP array as an actual parameter to the formal parameter and modifying it in the function will not affect the array itself, indicating that the transfer in this process is a value transfer, and the array variable is not. A reference to this array itself.
The operating environment of this tutorial: windows7 system, PHP version 8.1, DELL G3 computer
In PHP, most variables Types such as strings, integers, floating points, arrays, etc. are all value types, while classes and objects are reference types. When using them, you need to pay attention to this.
PHP variables are stored in memory like this. What the variables save is not directly the content of the value, but the address. For example:
$a = 1;
It seems to us that the variable $a directly stores the value 1. The actual situation is that the PHP interpreter creates the variable $a, stores the value: 1 somewhere in the memory, and then stores the address of the value in the variable $a.
When you need to get a value, first find the address in variable $a, and then find the value of the variable based on the address.
Look down:
echo $a;
will output 1. The PHP interpreter will complete this code like this: find the address stored in $a, and find the value stored somewhere in the memory based on the address. Output to screen.
It seems like a simple line of code, but the process is like this.
Look further down:
$a = 1; $b = &$a;
The variable $b here does an interesting operation. The & symbol takes out the address stored in the $a variable and stores it in the $b variable.
Then, if you use the following code:
echo $b;
The result will also be output 1. The PHP interpreter first takes out the address stored in $b, and then finds the value based on the address. If we do the following operations:
$a = 2; echo $b; //输出2
We will find that if the value of $a is changed, $b will also change. In fact, conversely, if the value of $b is changed, the value of $a will also change accordingly.
At this point, we can be sure that variables $a and $b store the same address and point to the same value.
Then, we can conclude that they represent the same variable.
To further summarize, it can be concluded that if two variables store the same address, they are the same variable.
After understanding some content, we begin to introduce value types and reference types.
Let’s look at the following code first:
$a = 1; $b = $a; $a = 2; echo $b; //输出 1
Assign the value of $a to $b. After changing the value of $a, the value of $b remains unchanged, that is Say $a and $b are two different variables pointing to different addresses. This form of assignment that creates different variables is called a value type.
Look again:
class User{ public $name = ‘Tome‘; } $a = new User; $b = $a; $a->name = ‘Jim‘; echo $b->name; //输出 Jim
Similarly assign the value of $a to $b. After changing $a, $b also changes, that is to say, $a and $b are the same variables, pointing to the same address. This form of assignment that does not create a new variable is called a reference type.
php array value type
Array passing in PHP is by value rather than by reference.
When calling a function, assign the PHP array as an actual parameter to the formal parameter, and modifying it in the function will not affect the array itself.
Explain that the transfer in this process is by value. The array variable is not a reference to the array itself. The PHP array itself exists in the form of a value, and the formal parameter is a copy of the array.
This is very different from other languages (such as c, Js, etc.), so it is worth noting!
The example is as follows:
$arr = array( 'name' => 'corn', 'age' => '24', ); test_arr($arr); function test_arr($arr){ $arr['name'] = 'qqyumidi'; } print_r($arr); //result: Array ( [name] => corn [age] => 24 )
js code is as follows:
var arr = new Array('corn', '24'); test_arr(arr); function test_arr(arr){ arr[0] = 'qqyumidi'; } console.log(arr); //result:["qqyumidi", "24"]
If you need to use the reference passing effect for value passing in PHP, you can add in front of the formal parameter Take the address character &.
$aa = 100; test_vars($aa); function test_vars(&$aa){ $aa = 200; } print_r($aa); //result: 200
Recommended learning: "PHP Video Tutorial"
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