What is the value of ~0 in C?

Question

I want to get the values of INT_MIN and INT_MAX. I've tried ~0 and ~0 >> 1 since the leftmost bit is a sign bit but I got -1 for both of them.

It's so confused that why ~0 doesn't turn out to be 0xffffffff and ~0 >> 1 to be 0x7fffffff?

Answer 1

Use:

~0U >> 1

Suffix 'U' for unsigned shift behavior.

so, confused that why not ~0 turns out to be 0xffffffff?

See, what is 0 say in four bytes representation:

BIT NUMBER    31                                     0
               ▼                                     ▼
number bits    0000 0000 0000 0000 0000 0000 0000 0000 
               ▲                                     ▲ 
              MSB                                   LSB



LSB - Least Significant Bit (numbered 0)
MSB - Most  Significant Bit (numbered 31) 

Now ~ is bitwise not operator then flips all bits in 0 as:

BIT NUMBER    31                                     0
               ▼                                     ▼
number bits    1111 1111 1111 1111 1111 1111 1111 1111
               ▲                                     ▲ 
              MSB                                   LSB

Because of MSB = 1 this representation is treated as negative number and its magnitude is find using 2'complement math that is -1.

How?

What is 1 ? it is:

number bits    0000 0000 0000 0000 0000 0000 0000 0001 
               ▲                                     ▲ 
              MSB                                   LSB

1's complement of 1

number bits    1111 1111 1111 1111 1111 1111 1111 1110
               ▲                                     ▲ 
              MSB                                   LSB

2'complement? Add 1 in one's complement, that is:

number bits    1111 1111 1111 1111 1111 1111 1111 1111
               ▲                                     ▲ 
              MSB                                   LSB

this same as when you gets ~0 ? that is why you are getting -1 output.

Now >> shift operator?

In most implementation of C >> operator is defined as an arithmetic right shift, which preserves the sign bit MSB. So ~0 >> 1 is noting but -1 remains same.

6.5.7 [Bitwise shift operators]

5 The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.

You requirement is what is called unsigned right shift >> and the behavior you needed can be find using unsigned number that is why I suffixed U as 0U.

How to print INT_MIN and INT_MAX?

Because printing INT_MIN and INT_MAX is bit tricky(because of undefined and implementation behavior of setting MSB and bit-overflow) in C so I have written a code as follows:

#include <stdio.h>
#include<limits.h> /* include for CHAR_BIT */
int main(){
  int my_int_min = 1U << ((sizeof(int) * CHAR_BIT) - 1);
  int my_int_max = ~0U >> 1;
  printf("INT_MIN  = %d\n", my_int_min);
  printf("INT_MAX  = %d\n", my_int_max);
  return 0;
}

See it executing @codepad, it output is:

INT_MIN  = -2147483648
INT_MAX  = 2147483647 

How does this code work?

Note for 32-bit number range is [-2147483648, 2147483647] that is equals to [-231, 231 -1 ].

INT_MIN: -2³¹ == -2147483648 is:

    1000 0000 0000 0000 0000 0000 0000 0000 
    ▲                                     ▲ 
    MSB                                   LSB

In expression 1U << ((sizeof(int) * CHAR_BIT) - 1), I shifts first bit the LSB(that is 1) to left most side at MSB, And because in C, setting signed bit is undefined behavior when operand is singed type so I used unsigned one 1U.

6.5.7 [Bitwise shift operators]

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

Another point to note is I used CHAR_BIT a standard macro defined in limits.h that tells number of bits in one char in a C implementation (remember: A char is always one byte size but number of bits in one bytes can be different on different system not always guaranteed to be 8).

INT_MAX: 2³¹ -1 == 2147483647

    0111 1111 1111 1111 1111 1111 1111 1111
    ▲                                     ▲ 
    MSB                                   LSB

Answer 2

0 is of type int. So are ~0 and ~0 >> 1 because of int type promotion

~0 has all 1s in its bit pattern and it's -1 in 2's complement, which is the default representation of most modern implementations.

Right shift in C is implementation defined. But most implementations define >> as arithmetic shift when the type is signed and logical shift when the type is unsigned

Since ~0 is int, which is a signed type, ~0 >> 1 will be an arithmetic shift right. Hence the value is sign extended, cause the value to be all 1s

You need to do unsigned(~0) >> 1 or ~0U

There are no ways to get INT_MIN and INT_MAX portably because in C there are 3 different signed type implementations beside trap representations and padding bits. That's why standard libraries always define INT_MIN and INT_MAX directly with the values

Answer 3

~0 is -1. Every C implementation that you're likely to run into uses two's complement for signed integers, so 0xffffffff is -1 (assuming 32-bit integers). ~0 >> 1 is the equivalent of dividing -1 by 2; since we're doing integer arithmetic, the result is -1.

Answer 4

The value of an all bit set int depends on the sign representation that your platform has for int. This is why the macros INT_MIN and INT_MAX were invented, there is no way of computing these values in a portable way.