直接上代码吧。希望可以得到大神的分析结果。
<code>$test = array(1, 2, 3, 4, 5); foreach ($test as &$value) { $value = 0; } unset($value); print_r($test); // 结果如下 Array ( [0] => 0 [1] => 0 [2] => 0 [3] => 0 [4] => 0 ) $test = array(1, 2, 3, 4, 5); foreach ((array)$test as &$value) { $value = 0; } unset($value); print_r($test); // 结果如下 Array ( [0] => 1 [1] => 2 [2] => 3 [3] => 4 [4] => 5 ) </code>
如上代码所示,仅仅是使用了一下数组强制类型转换,结果却是截然不同,我想知道为什么?
回复内容:
直接上代码吧。希望可以得到大神的分析结果。
<code>$test = array(1, 2, 3, 4, 5); foreach ($test as &$value) { $value = 0; } unset($value); print_r($test); // 结果如下 Array ( [0] => 0 [1] => 0 [2] => 0 [3] => 0 [4] => 0 ) $test = array(1, 2, 3, 4, 5); foreach ((array)$test as &$value) { $value = 0; } unset($value); print_r($test); // 结果如下 Array ( [0] => 1 [1] => 2 [2] => 3 [3] => 4 [4] => 5 ) </code>
如上代码所示,仅仅是使用了一下数组强制类型转换,结果却是截然不同,我想知道为什么?
先看这个例子
<code>$test = array(1, 2, 3, 4, 5); $test1 = $test; foreach ($test1 as &$value) { $value = 0; } print_r($test); Array ( [0] => 1 [1] => 2 [2] => 3 [3] => 4 [4] => 5 ) </code>
再看这个例子
<code>$test = array(1, 2, 3, 4, 5); $test1 = &$test; foreach ($test1 as &$value) { $value = 0; } print_r($test); Array ( [0] => 0 [1] => 0 [2] => 0 [3] => 0 [4] => 0 ) </code>
上面连个例子都是先赋值,然后再执行foreach,但是第一个是传值,而第二个是传引用(地址),传值相对于对原array 做了一个copy 赋值到新的变量上,他们在内存中指向不是同一个地方。
在看你的第二个例子中的 (array)$test, 这个其实相当于创建一个新的array,这个array 和test 肯定不是指向同一个内存地址,所以你foreach中的赋值操作并不会修改原来test变量中的值
(array)$test 这本身是一个赋值操作 等于 $a = (array)$test 然后再 foreach $a
数组(Array) 的赋值总是会涉及到值的拷贝。使用引用运算符&通过引用来拷贝数组
<code>php</code><code>$test = array(1, 2, 3, 4, 5); $test2 = (array)$test; $test2[0] = 0; print_r($test);//[1,2,3,4,5] print_r($test2);//[0,2,3,4,5] </code>
foreach遍历数组的时候,不加引用他每遍历一次都会释放掉value得值,如果加了引用,他不会释放,而被后来的给覆盖掉[图片]要相信我,
两种代码我输出opcode,来针对opcode做讨论.
在做foreach(xxx as xx)这类操作的时候,调用的指令是FE_REST来重置数组的迭代器.
从我截图的红框里可以清楚的看到,第一次重置的时候用的是!0,也就是$test,第二次用的是~2,也就是(array)$test.
所以你第一次更改的时候,改变的是$test里的内容,而第二次更改的时候,改变的是(array)$test这个变量里的内容,意思就是第二次根本就没更改到$test

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