To define a function that accepts referenced parameters in PHP, you need to prepend the parameter with &: function increment(&$number) { $number ; } when defining the function. 1. When defining a function, add the & symbol before the parameter name to enable reference passing; 2. When calling the function, do not need to add &, just pass in the variable directly; 3. Do not use reference passing on the literal, otherwise an error will be reported; 4. Reference passing is suitable for situations where external variables need to be modified, but abuse should be avoided to keep the code clear; 5. PHP also supports returning references, but should be used with caution. For example, after calling increment($num), the value of $num will be modified internally by the function and preserved externally.
In PHP, by default, the parameters of the function are passed by value, which means that the modification of the variables inside the function will not affect the original variables outside. But if you want the function to modify the variables passed in and these modifications can be reflected outside the function, you need to use pass by reference .

PHP allows you to specify that the parameter is passed in reference by preceding the parameter by specifying the &
symbol when defining a function.

How to define a function that accepts referenced parameters?
When defining a function, add &
before the parameter name:
function increment(&$number) { $number ; }
When called in this way, the variables passed in will be treated as references:

$num = 5; increment($num); echo $num; // Output 6
As you can see, the value of $num
is changed in the function, and this change is retained outside the function.
What should be noted is:
- Reference passes must be declared when the function is defined and cannot be temporarily decided upon when the call is called.
- It is not recommended to pass references to literals (such as numbers and direct strings) as an error will be triggered.
How to correctly use reference parameters when calling a function
When calling a function, you do not need to add &
before the parameters, you only need to make sure that the reference parameters have been declared in the function definition.
$value = 10; increment($value); // Correct: no need to add & when calling
But if you see a writing method like this in the old version of PHP code:
increment(&$value); // Obsolete writing
This was intended to explicitly referencing passes in earlier versions, but was no longer needed in PHP 5 and later, and this writing is now considered outdated and may even report an error.
Common misunderstandings and precautions
Not all parameters need to be passed by reference
If you just want to read the value of a variable instead of modifying it, there is no need to use a reference. Abuse of references will make the code difficult to understand and maintain.Will reference passes affect performance?
Generally speaking, reference passes can save memory, especially when dealing with large arrays or objects. However, PHP has been optimized internally (such as copy-on-write), so in most cases, there is no need to deliberately use references to improve performance.Returning a quote is also feasible
In addition to parameters, PHP also supports returning references, but this is more complicated and you need to be particularly careful when using it. For example:function &getReference() { static $var = 0; return $var; } $ref = &getReference(); $ref = 10; echo getReference(); // Output 10
Summarize
The key to implementing reference parameter transfer in PHP is to add
&
to the parameter when the function is defined. This method is suitable for situations where you need to modify external variables inside the function. Be careful not to abuse it, keeping the code clear and secure is more important.Basically that's it.
The above is the detailed content of How to pass arguments by reference in a PHP function?. For more information, please follow other related articles on the PHP Chinese website!

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