Home >Backend Development >PHP Tutorial >Regarding the problem of & pass-by-value reference in PHP, can you help explain the principle of the output result in the result of the foreach loop? ,PHP_PHP tutorial
Regarding the problem of & pass-by-value reference in PHP, can you help explain the principle of the output result in the result of the foreach loop? ,PHP_PHP tutorial
- WBOYOriginal
- 2016-07-12 08:54:00772browse
Regarding the problem of & pass-by-value reference in PHP, can you help explain the principle of the output result in the result of the foreach loop? , PHP
PHP中的&传值引用的问题，在foreach循环的结果能帮解释下输出的结果原理是什么？ <br />代码如下： <br /><?php <br />$arr = array('one','two','three'); <br />foreach ($arr as &$value){ echo 'Value:'.$value.'<br />'; } <br />foreach ($arr as $value){ echo 'Value:'.$value.'<br />'; } <br />?><br />输出结果：<br /> Value:one<br /> Value:two <br /> Value:three <br /><br /> Value:one <br /> Value:two <br /> Value:two
<br />第一次带&的foreach并没有改变数组的内容。。<br />而是最后一次循环$value引用了数组的最后一个项 (可以测试一下，在第一次循环结束后unset($value)，第二次循环的结果就不会有变化)，<br />在你第二个foreach也是使用的$value变量，这才造成了怪异的问题(可以换个变量,比如$val，输出的数组就不会有变化)。<br />第二个foreach是赋值给$value，但是这时的$value是引用的数组的最后一个值，<br />所以<br />第一次循环把one赋值给了最后一个值，<br />第二次把two赋值给最后一个，<br />第三次也就是最后一个已经在第二次循环被赋值为two，所以仍然是two。
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