c 中的 printf 波浪号运算符

Question

I know that the ~ operator is NOT, so it inverts the bits in a binary number

unsigned int a = ~0, b = ~7;
printf("%d\n",a);
printf("%d\n",b);
printf("%u\n",a);
printf("%u\n",b);

I guessed 0 will be 1 and 7 (0111) will be 8 (1000) but the output was

-1
-8
4294967295
4294967288

how did ~0 and ~7 become -1, and -8? also why is %u printing that long number?

Answer 1

The ~ operator simply inverts all bits in a number.

On most modern compilers, int is 32 bits in size, and a signed int uses 2's complement representation. Which means, among other things, that the high bit is reserved for the sign , and if that bit is 1 then the number is negative.

0 and 7 are int literals. Assuming the above, we get these results:

• 0 is bits 00000000000000000000000000000000b
  = 0 when interpreted as either signed int or unsigned int

• ~0 is bits 11111111111111111111111111111111b
  = -1 when interpreted as signed int
  = 4294967285 when interpreted as unsigned int

• 7 is bits 00000000000000000000000000000111b
  = 7 when interpreted as either signed int or unsigned int

• ~7 is bits 11111111111111111111111111111000b
  = -8 when interpreted as signed int
  = 4294967288 when interpreted as unsigned int

In your printf() statements, %d interprets its input as a signed int , and %u interprets as an unsigned int . This is why you are seeing the results you get.

Answer 2

The ~ operator inverts all bits of the integer operand. So for example where int is 32-bit, 1 is 0x00000001 in hex and it's one's complement is 0xFFFFFFFE. When interpreted as unsigned, that is 4 294 967 294, and as two's complement signed, -2.