c++ - 有符号和无符号的位移位操作

Question

我正在对一个 int 进行移位操作，结果没有按预期进行，这让我感到很惊讶。

int i, res;
i = 0x80000000;
res = i>>1;                //results in 0xc0000000
res = (unsigned int) i>>1; //results in 0x40000000

整数中的一位移位怎么可能只对第 31 位起作用？

Answer 1

你看到的可能是arithmetic bit shift .

when shifting to the right, the leftmost bit (usually the sign bit in signed integer representations) is replicated to fill in all the vacant positions (this is a kind of sign extension).

C99 standard 6.5.7§5 说:

The result of E1 >> E2 is E1 right-shifted E2 bit positions. [...] If E1 has a signed type and a negative value, the resulting value is implementation-defined.

因此，结果可能是编译器编写者决定的任何结果。他们可能决定扩展符号位，编译器文档应该提到它。