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php中函数的形参与实参的问题说明_php技巧

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Release： 2016-05-17 09:23:18
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当实参个数形参个数 时，php是不会报错的，它只会取前面的几个参数，多余的则将会丢弃。

在PHP中编写函数,一般情况下调用函数的时候,改变的值都是形参而不是实参.但是如果在形参中加入地址符时候就会改变实参的值,为什么?

请看下面的例子:

复制代码 代码如下:

//编写一个函数swap(),测试该函数的实参值无改变
function swap($a,$b) {
echo "

进入swqp()函数前
\n";
echo "交换前:形参a=$a,形参b=$b
\n";
$c=$b;
$a=$b;
$b=$c;
echo "交换后:形参a=$a,形参b=$b
\n";
echo "退出swap()函数

\n";
}
$variablea=5;
$variableb=10;
echo "调用swap()函数前: ";
echo "实参a=$variablea,实参b=$variableb
\n";
swap($variablea,$variableb);
echo "调用swap()函数后: ";
echo "实参a=$variablea,实参b=$variableb
\n";
?>

复制代码 代码如下:

//测试swap()函数实参的值改变
function swap1(&$a,&$b) {
echo "

进入swap1()函数
\n";
echo "交换前: 形参a=$a,形参b=$b
\n";
$c=$b;
$a=$b;
$b=$c;
echo "交换后: 形参a=$a,形参b=$b
\n";
echo "退出swap()函数

\n";
}

$variablea=5;
$variableb=10;
echo "调用swap1()函数前: ";
echo "实参a=$variablea,实参b=$variableb
\n";
swap1($variablea,$variableb);
echo "调用swap1()函数后: ";
echo "实参a=$variablea,实参b=$variableb
\n";
?>

//上面的两个例子就是说明,求教了~~~
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