试图理解 openssl_sign 中的 &$signature 参数

Question

I'm trying to understand -> https://www.php.net/manual/en/function.openssl-sign.php which has &$signature parameter. I read this -> PHP &$string - What does this mean? but still don't understand why must we have &$signature as a parameter or how do we use it. The examples at https://www.php.net did not describe the use of that variable.

Here's one of their examples.

<?php
// $data is assumed to contain the data to be signed

// fetch private key from file and ready it
$pkeyid = openssl_pkey_get_private("file://src/openssl-0.9.6/demos/sign/key.pem");

// compute signature
openssl_sign($data, $signature, $pkeyid);

// free the key from memory
openssl_free_key($pkeyid);
?>

Btw, have a fabulous coding new year 2022!

Answer 1

It means you're passing a variable 'by reference' ( https://www.php.net/manual/en/language.references.pass.php ), which, in the simplest terms, means that the variable you pass can be modified by the method or function it's passed to without having to return it.

For example:

$a = 1;

function foo($value) {
    $value++;
}

foo($a);

// $a is still 1
var_dump($a);

Whereas specifying the function's signature as &$value :

$b = 1;

function bar(&$value) {
    $value++;
}

bar($b);

// $b is 2
var_dump($b);

So the openssl_sign() function is likely to be modifying the value of $signature without you having to request it's updated value via a getter.

It should also be noted that objects are always passed by reference, so do not need the &$ signature - their state will always be modified and can be used outside of the function they are passed to subsequently.

Answer 2

Function 'openssl_sign' is defined to computes a signature which is passed by reference.

Without '&', parameter signature won't change after the function is called.

Answer 3

If you pass vars by reference to a function so the function can modify the variable directly without returning value.

https://www.php.net/manual/en/language.references.pass.php