c - C中的指针递增

Question

考虑以下代码:

unsigned short i;
unsigned short a;
unsigned char *pInput = (unsigned char *)&i;

pInput[0] = 0xDE;
pInput[1] = 0x01;

a = ((unsigned short)(*pInput++)) << 8 | ((unsigned short)(*pInput++));

为什么a的值为0xDEDE，而不是0xDE01？

Answer 1

代码调用了未定义的行为。原因是 pInput 在两个 sequence points 之间修改了不止一次.您可能会得到任何预期或意外的结果。无话可说。

C99 规定:

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored.

阅读c-faq 3.8以获得更详细的解释。