What does & before the function name signify?

What does the & before the function name signify?

Does that mean that the $result is returned by reference rather than by value? If yes then is it correct? As I remember you cannot return a reference to a local variable as it vanish开发者_如何学Ces once the function exits.

function &query($sql) {
 // ...
 $result = mysql_query($sql);
 return $result;
}

Also where does such a syntax get used in practice ?

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Does that mean that the $result is returned by reference rather than by value?

Yes.

Also where does such a syntax get used in practice ?

This is more prevalent in PHP 4 scripts where objects were passed around by value by default.

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To answer the second part of your question, here a place there I had to use it: Magic getters!

class FooBar {
    private $properties = array();

    public function &__get($name) {
        return $this->properties[$name];
    }

     public function __set($name, $value) {
        $this->properties[$name] = $value;
    }
}

If I hadn't used & there, this wouldn't be possible:

$foobar = new FooBar;
$foobar->subArray = array();
$foobar->subArray['FooBar'] = 'Hallo World!';

Instead PHP would thrown an error saying something like 'cannot indirectly modify overloaded property'.

Okay, this is probably only a hack to get round some maldesign in PHP, but it's still useful.

But honestly, I can't think right now of another example. But I bet there are some rare use cases...

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Does that mean that the $result is returned by reference rather than by value?

No. The difference is that it can be returned by reference. For instance:

<?php
function &a(&$c) {
    return $c;
}
$c = 1;
$d = a($c);
$d++;
echo $c; //echoes 1, not 2!

To return by reference you'd have to do:

<?php
function &a(&$c) {
    return $c;
}
$c = 1;
$d = &a($c);
$d++;
echo $c; //echoes 2

Also where does such a syntax get used in practice ?

In practice, you use whenever you want the caller of your function to manipulate data that is owned by the callee without telling him. This is rarely used because it's a violation of encapsulation – you could set the returned reference to any value you want; the callee won't be able to validate it.

nikic gives a great example of when this is used in practice.

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  <?php
    // You may have wondered how a PHP function defined as below behaves:
    function &config_byref()
    {
        static $var = "hello";
        return $var;
    }
    // the value we get is "hello"
    $byref_initial = config_byref();
    // let's change the value
    $byref_initial = "world";
    // Let's get the value again and see
    echo "Byref, new value: " . config_byref() . "\n"; // We still get "hello"
    // However, let’s make a small change:
    // We’ve added an ampersand to the function call as well. In this case, the function returns "world", which is the new value.
    // the value we get is "hello"
    $byref_initial = &config_byref();
    // let's change the value
    $byref_initial = "world";
    // Let's get the value again and see
    echo "Byref, new value: " . config_byref() . "\n"; // We now get "world"
    // If you define the function without the ampersand, like follows:
    // function config_byref()
    // {
    //     static $var = "hello";
    //     return $var;
    // }
    // Then both the test cases that we had previously would return "hello", regardless of whether you put ampersand in the function call or not.