php - 为什么这段代码不简单地打印字母 A 到 Z？

Question

<?php
for ($i = 'a'; $i <= 'z'; $i++)
    echo "$i\n";

此代码段给出以下输出(换行符被空格替换):

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap aq ar as at au av aw ax ay az ba bb bc bd be bf bg bh bi bj bk bl bm bn bo bp bq br bs bt bu bv bw bx by bz ca cb cc cd ce cf cg ch ci cj ck cl cm cn co cp cq cr cs ct cu cv cw cx cy cz da db dc dd de df dg dh di dj dk dl dm dn do dp dq dr ds dt du dv dw dx dy dz ea eb ec ed ee ef eg eh ei ej ek el em en eo ep eq er es et eu ev ew ex... on to yz

Answer 1

来自docs :

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.

From Comments:-
还应注意 <=是一个字典比较，所以 'z'+1 ≤ 'z' . (从 'z'+1 = 'aa' ≤ 'z' 开始。但是 'za' ≤ 'z' 是第一次比较为假。)在 $i == 'z' 时中断例如，会起作用。

Example here .