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函数返回引用解决方法
- WBOYOriginal
- 2016-06-13 12:20:21790browse
函数返回引用
function &bar()
{
$a = 5;
return $a;
}
$c=bar();
echo $c;
上面的函数不是运行完毕 函数体的变量不是给释放了 怎么返回的引用还有效呢？
------解决思路----------------------
运行结果不是5么？你return 返回给$c了
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