php变量前面加&符号是什么意思就不用多说了,大家都在用,就是两个变量同时指向一个地址而已,那么,php函数前面加&符号的意义又是什么呢?下面先上两个演示代码,然后再做解释。
function &chhua() { static $b="www.jb51.net";//申明一个静态变量 $b=$b."WEB开发"; echo $b; return $b; } $a=chhua();//这条语句会输出 $b的值 为“www.jb51.netWEB开发” $a="PHP"; echo "<Br>"; $a=chhua();//这条语 句会输出 $b的值 为“www.jb51.netWEB开发WEB开发” echo "<Br>"; $a=&chhua();//这条语句会输出 $b的值 为“www.jb51.netWEB开发WEB开发WEB开发” echo "<Br>"; $a="JS"; $a=chhua(); //这条语句会输出 $b的值 为"JSWEB开发" function &test() { static $b=0;//申明一个静态变量 $b=$b+1; echo $b; return $b; } $a=test();//这条语句会输出 $b的值 为1 $a=5; $a=test();//这 条语句会输出 $b的值 为2 $a=&test();//这条语句会输出 $b的值 为3 $a=5; $a=test(); //这条语句会输出 $b的值 为6
下面就来解释一下第二个函数吧。
通过这种方式$a=test();得到的其实不是函数的引用返回,这跟普通的函数 调用没有区别.
至于原因: 这是PHP的规定
php规定通过$a=&test(); 方式得到的才是函数的引用返回.
至于什么是引用返回呢(PHP手册上说:引用返回用在当想用函数找到引用应该被绑定在哪一个变量上面时。)
用上面的例子来解释就是
$a=test()方式调用函数,只是将函数的值赋给$a而已, 而$a做任何改 变 都不会影响到函数中的$b.
而通过$a=&test()方式调用函数呢, 他的作用是将return $b中的 $b变量的内存地址与$a变量的内存地址,
指向了同一个地方. 即产生了相当于这样的效果($a=&b;) 所以改变$a的值,也同时改变了$b 的值 ,所以在执行了:
$a=&test(); $a=5; 以后,$b的值变为了5。

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