PHP对字符串的递增运算分析
Jun 13, 2016 pm 12:14 PM
echoforphpcodeanalyzecopystringknowledgerightofOperationIncrementquestion
有同学问了一个问题:
复制代码 代码如下:
for($i = 'A'; $i echo $i;
}
输出是啥?
输出是:
ABCDEFGHIJKLMNOPQRSTUVWXYZAAABACADAEAFAGAHAIAJAKALAMANAOAPAQARAS…….
为啥?
其实很简单, PHP的手册中也有说明, 只不过恐怕很多人不会一章一节的把手册仔细阅读一遍:
PHP follows Perl's convention when dealing with arithmetic operations on character variables and not C's. For example, in Perl ‘Z'+1 turns into ‘AA', while in C ‘Z'+1 turns into ‘[‘ ( ord(‘Z') == 90, ord(‘[‘) == 91 ). Note that character variables can be incremented but not decremented and even so only plain ASCII characters (a-z and A-Z) are supported.
在处理字符变量的算数运算时,PHP 沿袭了 Perl 的习惯,而非 C 的。例如,在 Perl 中 ‘Z'+1 将得到 ‘AA',而在 C 中,'Z'+1 将得到 ‘[‘(ord(‘Z') == 90,ord(‘[‘) == 91)。注意字符变量只能递增,不能递减,并且只支持纯字母(a-z 和 A-Z)。
也就是说, 如果:
复制代码 代码如下:
$name = "laruence";
++$name; //将会是"laruencf"
而:
复制代码 代码如下:
$name = "laruence";
--$name; //没有影响, 还是"laruence"
所以, 这个问题的原因就是当$i = Z的时候, ++$i成了AA, 而字符串比较的话,
AA,BB,XX一直到YZ都是小于等于Z的… so..
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