[英]Bitwise operators on signed ints
I was curious why the following only works when defining an unsigned
char:我很好奇为什么以下仅在定义
unsigned
字符时才有效:
#define BITS 8
unsigned char d = 0b00001011; // will fail if doing `char d`
d = ~d;
char buffer[BITS+1] = "00000000";
for(int ix=0; d!=0; d>>=1, ix++) {
buffer[BITS-1-ix] = d&1 ? '1' : '0';
}
printf("%s\n", buffer);
Otherwise I get a SegFault, which I'm guessing is due to the d>>=1
on the signed type.否则我会得到一个 SegFault,我猜这是由于签名类型上的
d>>=1
造成的。 Why does that occur exactly though?为什么会发生这种情况呢? Wouldn't it have the same bit pattern and doing
>>1
would just push the bits to the right once?它不会具有相同的位模式并且执行
>>1
只会将位向右推一次吗?
Shifting a negative signed number rightward has implementation-defined behaviour;将负符号数右移具有实现定义的行为; exactly what it does will depend on your compiler.
它的确切作用取决于您的编译器。
Likely your compiler is using two's complement and arithmetic shifts right, ie the sign bit is filled with a copy of the bit that left it upon every shift.您的编译器可能正在使用二进制补码和算术右移,即符号位填充有每次移位时留下的位的副本。
This is often a better choice because eg it means that -4 >> 1
is -2
rather than, in an 8-bit quantity, being 126
.这通常是一个更好的选择,因为例如它意味着
-4 >> 1
是-2
而不是 8 位数量的126
。
Slightly off topic, but the simplest fix for your code is just to switch the exit condition to d != 0 && ix < 8
.稍微偏离主题,但对您的代码最简单的解决方法是将退出条件切换为
d != 0 && ix < 8
。
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