在C ++中增加char指针

Question

为什么这个节目，

char *s, *p, c;

s = "abc";

printf(" Element 1 pointed to by S is '%c'\n", *s);
printf(" Element 2 pointed to by S is '%c'\n", *s+1);
printf(" Element 3 pointed to by S is '%c'\n", *s+2);
printf(" Element 4 pointed to by S is '%c'\n", *s+3);
printf(" Element 5 pointed to by S is '%c'\n", s[3]);
printf(" Element 4 pointed to by S is '%c'\n", *s+4);

给出以下结果？

 Element 1 pointed to by S is 'a'
 Element 2 pointed to by S is 'b'
 Element 3 pointed to by S is 'c'
 Element 4 pointed to by S is 'd'
 Element 5 pointed to by S is ' '
 Element 4 pointed to by S is 'e'

编译器是如何继续序列的？ 为什么s[3]返回一个空值？

Answer 1

它没有继续序列。 你正在做*s+3 ，它首先解引用s给你一个值为'a'的char ，然后你添加到那个char值。 将3添加到'a'会给出'd'的值（至少在执行字符集中）。

如果将它们更改为*(s+1) ，依此类推，您将获得预期的未定义行为。

s[3]访问字符串的最后一个元素，即空字符。

Answer 2

请注意， *s是一个字符，本质上是一个数字。 添加另一个数字会导致ASCII值更高的字符。 s[3]为空，因为您只将“abc”分别分配给条目0,1,2。 实际上第三个字符是'\\ 0'字符。