C 语言中的位取反操作符
话不多说，先来看一段代码：
1 #include <stdio.h> 2 3 int main(int argc, char *argv[]) 4 { 5 unsigned char uc = 0x1f; 6 unsigned int ui = ~uc; 7 unsigned long long ull = ~uc; 8 9 unsigned char cc = ~uc; 10 printf("uc = %#x\n", uc); 11 printf("cc: ~uc = %#x\n", cc); 12 printf("ui: ~uc = %#x\n", ui); 13 printf("ull: ~uc = %#llx\n", ull); 14 return 0; 15 }
因为用到了 C99 标准的 long long 类型，所以编译时候需要添加 "-std=c99" 选项：
gcc -Wall -std=c99 -o file file.c
输出结果(x86 mingw32)：
uc = 0x1f
cc: ~uc = 0xe0
ui: ~uc = 0xffffffe0
ull: ~uc = 0xffffffffffffffe0
这里显现出了 位取反 操作的结果特殊性，对于通常而言，我们以为一个 unsigned char 位取结果的类型也应该是 unsigned char ，也就是说当我们把这个结果赋给一个 unsigned int （或者 unsigned long long）类型的值时候，这个值也应该是 不会超过 unsigned char 的取值范围的。但是上述代码显示出并非这样的情况。
上述代码显示出把 unsigned char 位取反值赋给 unsigned int 变量时，除了低位字节是补码外，高位字节都是 0 的补码 1了。对于赋值给 unsigned long long 变量也是如此。
我个人理解的解释是，在 x86-32 的机器上，机器字长为 32，寄存器长度也是32，对于像 unsigned char 这样的 1 byte 的值，运算时会存储在寄存器中的低位字节，高位字节都为0，然后位取反操作指令会对整个寄存器的值位取反，然后赋值给 unsigned int (4 bytes)这样的变量时，是将整个寄存器值取出。这样高位字节都会是 0 的补码 1。但是这样怎么解释赋值给 unsigned long long 类型变量发生的情况呢？或者说对于位取反操作的结果是个特殊状态的值？
PS: 最近在看 CSAPP（《深入理解计算机系统》）第二版，其中在做练习题 2.12 发现了上述情形，之前不知道，所以就研究了一下下，不过还是半解。。。。
Apr 24, 2012 08:45:35 PM
我查了一下，根据 c99 标准 6.3.1.1 节的说明，似乎对于级别比 int 和 unsigned int 低的整型变量，只要其范围是 int 的子区间，都会隐式强转为 int 来参与运算。
"If an int can represent all values of the original type, the value is converted to an int;
otherwise, it is converted to an unsigned int. These are called the integer
promotions.48) All other types are unchanged by the integer promotions.
"
Apr 24, 2012 08:48:29 PM
http://learn.akae.cn/media/ch15s03.html
这里有该特性的中文介绍和使用示例。
Apr 25, 2012 08:35:37 AM
@vx13: 但是 对于 转换成 long long 那里的区别还是一知半解状态
Apr 25, 2012 09:12:50 AM
@tisyang: 那个是赋值的问题。把 int 型的 -1 赋给 unsigned long long 的变量，就会变成一堆 f 的。
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