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What References Are Not

As said before, references are not pointers. That means, the following construct won't do what you expect:

代码语言:javascript
复制
<?php
function foo(&$var)
{
    $var =& $GLOBALS["baz"];
}
foo($bar); 
?>

What happens is that $var in foo will be bound with $bar in the caller, but then re-bound with $GLOBALS["baz"]. There's no way to bind $bar in the calling scope to something else using the reference mechanism, since $bar is not available in the function foo (it is represented by $var, but $var has only variable contents and not name-to-value binding in the calling symbol table). You can use returning references to reference variables selected by the function.

← What References Do

Passing by Reference →

代码语言:txt
复制
 © 1997–2017 The PHP Documentation Group

Licensed under the Creative Commons Attribution License v3.0 or later.

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