为什么此代码片段的输出为 -33

#include<stdio.h>

int main()
{
    int a=32;
    printf("%d\n", ~a);  //line 2
    return 0;
}

o/p = -33

实际上在原始片段第 2 行中是

 printf("%x\n", ~a);  //line 2

我解决了它像

32 in hex is 20.
0000 0000 0010 0000
now tilde operator complements it
1111 1111 1101 1111 = ffdf.

当我有时我很困惑如何解决它

printf("%d\n", ~a);  //line 2 i.e %d NOT %x.

在 C 实现中，与任何编程语言的大多数现代实现一样，有符号整数用补码 http://en.wikipedia.org/wiki/Two%27s_complement.

在二进制补码中，高位表示负数，并且值的编码方式如以下示例所示：

Bits  Decimal
0…011 +3
0…010 +2
0…001 +1
0…000  0
1…111 -1
1…110 -2
1…101 -3

Thus, if the usual (unsigned) binary value for the bits is n and the high bit is zero, the represented value is +n. However, if the high bit is one, then the represented value is n-2^w, where w is the width (the number of bits in the format).

So, in an unsigned 32-bit format, 32 one bits would normally be 4,294,967,295. In a two’s complement 32-bit format, 32 one bits is 4,294,967,295 - 2³² = -1.

In your case, the bits you have are 1111 1111 1111 1111 1111 1111 1101 1111. In unsigned 32-bit format, that is 4,294,967,263. In two’s complement, it is 4,294,967,263 - 2³² = -33.