C Boolean – What is !0 in C?

This is just an initializer list for an array. So it's very like the normal syntax:

char buf[5] = { 1, 2, 3, 4, 5 };

However, the C standard states that if you don't provide enough elements in your initializer list, it will default-initialize the rest of them. So in your code, all elements of buf will end up initialized to 0.

printf doesn't display anything because buf is effectively a zero-length string.

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