为什么“回声'2'. print(2)+ 3“；打印521? [英] Why does "echo '2' . print(2) + 3" print 521?

问题描述

有人可以告诉下面表达式的内部过程吗?

Can anybody tell the internal procedure of the below expression?

<?php echo '2' . print(2) + 3; ?>
// outputs 521

推荐答案

回显由以下内容组成的串联字符串:

Echo a concatenated string composed of:

字符串"2" 函数print('2')的结果，该结果将返回true，并将其字符串化为1 字符串"3"

The string '2' The result of the function print('2'), which will return true, which gets stringified to 1 The string '3'

现在，操作顺序在这里真的很有趣，根本不能以521结尾！让我们尝试一种变体，以找出问题所在.

Now, the order of operations is really funny here, that can't end up with 521 at all! Let's try a variant to figure out what's going wrong.

echo'2'.print(2)+ 3; 这将产生521

echo '2'.print(2) + 3; This yields 521

PHP正在将其解析为:

PHP is parsing that, then, as:

echo'2'. (print('2')+'3')) 答对了！左边的打印首先被评估，打印为"5"，这使我们离开

echo '2' . (print('2') + '3')) Bingo! The print on the left get evaluated first, printing '5', which leaves us

echo'1'.打印('2') 然后对左侧的图片进行评估，因此我们现在打印了"52"，剩下的是

echo '1' . print('2') Then the left print gets evaluated, so we've now printed '52', leaving us with

echo'1'. '1'; 成功. 521.

echo '1' . '1' ; Success. 521.

我强烈建议不要回显打印结果，也不要打印回显结果.首先，这样做是毫无意义的.

I would highly suggest not echoing the result of a print, nor printing the results of an echo. Doing so is highly nonsensical to begin with.

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