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通过5个php实例细致说明传值与传引用的区别

Jun 13, 2016 am 11:59 AM
php and learn Pass value the difference principle Example what is use want illustrate pass

哈哈,会用只是初级阶段,要了解原理是什么,这样才能更好去运用,费话不多说
传值:是把实参的值赋值给行参 ,那么对行参的修改,不会影响实参的值
传引用 :真正的以地址的方式传递参数传递以后,行参和实参都是同一个对象,只是他们名字不同而已对行参的修改将影响实参的值
说明:
传值:根copy是一样的。打个比方,我有一橦房子,我给你建筑材料,你建了一个根我的房子一模一样的房子,你在你的房子做什么事都不会影响到我,我在我的房子里做什么事也不会影响到你,彼此独立。
传引用:让我想起了上大学时学习C语言的指针了,感觉差不多。打个比方,我有一橦房子,我给你一把钥匙,我们二个都可以进入这个房子,你在房子做什么都会影响到我。
一,php实例
1,传值

复制代码 代码如下:


$param1=1; //定义变量1
$param2=2; //定义变量2
$param2 = $param1; //变量1赋值给变量2
echo $param2; //显示为1
?>


2,传引用

复制代码 代码如下:


$param2=1; //定义变量2
$param1 = &$param2; //将变量2的引用传给变量1
echo $param2; //显示为1
$param1 = 2; //把2赋值给变量1
echo $param2; //显示为2
?>


3,函数传值

复制代码 代码如下:


//传值
$param1 = 1; //定义变量1
function add($param2) //传参数
{
$param2=3; //把3赋值给变量2
}
$param3=add($param1); //调用方法add,并将变量1传给变量2
echo '
$param1=='.$param1.'
'; //显示为$param1==1
echo '
$param2=='.$param2.'
'; //显示为$param2== 因为$param2是局部变量,所以不能影响全局
echo '
$param3=='.$param3.'
'; //显示为$param3== 因为add方法没有返回值,所以$param3为空
?>


4,函数传引用

复制代码 代码如下:


//传值
$param1 = 1; //定义变量1
function add(&$param2) //传参数
{
$param2=3; //把3赋值给变量2
// return $param2; //返回变量2
}
echo '
$param1=='.$param1.'
'; //显示为$param1==1 没对变量1进行操作
$param3=add($param1); //调用方法add,并将变量1的引用传给变量2
echo '
$param1=='.$param1.'
'; //显示为$param1==3 调用变量过程中,$param2的改变影响变量1,虽然没有return
echo '
$param2=='.$param2.'
'; //显示为$param2== 因为$param2局部变量,所以不能影响全局
echo '
$param3=='.$param3.'
'; //显示为$param3== 如果把方法里面的return注释去掉的话就为$param3==3
?>


5,函数传引用2

复制代码 代码如下:


//传引用
$param1 = 1;
function &add(&$param2)
{
$param2 = 2;
return $param2;
}
$param3=&add($param1);
$param4=add($param1);
echo '
$param3=='.$param3.'
'; //显示为$param3==2
echo '
$param4=='.$param4.'
'; //显示为$param4==2
echo '
$param1=='.$param1.'
'; //显示为$param1==2 调用变量过程中,$param2的改变影响变量1
$param3++;
/*下面显示为$param1==3,这是因为$param2和$param1引用到同一个地方,
* 返回值前面加了地址符号还是一个引用$param3=&add($param1);
* 这样$param3,$param2和$param1引用到同一个地方,当$param3++;时,
* $param1会被改变*/
echo '
$param1=='.$param1.'
';
$param4++;
/* 下面显示为$param1==3,这里为什么是3而不是4呢,这是因为返回值前面没有
* 地址符号,它不是一个引用所以当$param4改变时不会影响$param1*/
echo '
$param1=='.$param1.'
';
?>


哈哈,不过我觉得传引用会好一点,耗的资源少。上面测试没有明显的差距,可能是因为测试数据不够大造成的,如果有更大数据来测试,我想会有明显的不同。
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