Clarification of PHP manual; default values passed by reference

问题:

Due to inconsistencies in the PHP manual (as I've posted about before) I'm just inquiring about some clarification.

The Function Arguments page (http://ca2.php.net/manual/en/functions.arguments.php) has the following note:

Note: As of PHP 5, default values may be passed by reference.

Now, I assume this simply means that the following syntax is acceptable:

function foo(&$bar = null){     // ... } 

However, again due to other inconsistencies, I was wondering if perhaps this pertains to something else.

回答1:

It means that in PHP 4, using a default value for arguments passed by reference would result in a parse error:

Parse error: syntax error, unexpected '=', expecting ')' in ... 

Demo

In PHP5, when no argument is passed, your function will have a normal local variable called $bar initialized to null.

It should probably be reworded to:

Note: As of PHP 5, function declarations may define a default value for argument passed by reference.

回答2:

it means that when you change bar

$bar = "newvalue"; 

in function, old (original one) will be affected too

<?php function foo(&$bar = null){     $bar = 'newval'; }  $bar = 'oldval, will be changed'; foo($bar); echo $bar; //RETURNS newval 

so if you change any variable passed by reference, it doesn't matter where you changed, source one is changed, too

http://sandbox.phpcode.eu/g/51723

回答3:

I think the only reason this exist is for allowing to skip trailing parameters in a function call

function test(&$bar = 10) {     echo " '$bar' ";     $bar = $bar*2;     echo " '$bar' "; }  test($aaa);     // prints '' '0'     (NULL as string, NULL*2) echo $aaa;      // prints 0          ($aaa is set to NULL*2) echo "<br>";  $bbb = 6; test($bbb);     // prints '6' '12'   (6, 6*2) echo $bbb;      // prints 12         ($bbb is set to 6*2) echo "<br>";  test();        // prints '10' '20' // (uses the default value since the argument was skipped, so: 10, 10*2) 

So imho the reason this exist is merely the possibility to have $bar set to some default value inside the function scope when you skip the leading parameter in the function's call

If it's like so I agree with you, the manual should be more precise about this

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