Why am I getting a wrong binary pattern?

I'm writing a function to print the Binary Representation of an integer which is 4-Bytes on my machine. I'm creating a mask = 0x80000000 which should be ( 10000000 00000000 00000000 00000000 )

and creating a loop to shift this mask right and making and operation and if it is one I'll print 1 else I'll print 0, but I'm getting wrong results!

For example: for decimal 10 I'm getting 00000000 00000000 00000000 00001111.

#include <stdio.h>

void Binary( int ) ;
int main()
{
    Binary(10) ; // 00000000 00000000 00000000 00001111
    while(1);
    return 0;
}

void Binary( int num )
{
    int mask = 0x80000000 ;
    for(int i = 0 ; i<32; i++)
    {
        if( (mask>>i)&num )
        {
            printf("1") ;
        }
        else
        {
            printf("0") ;
        }
    }
}

When you shift the int right, it fills with the sign bit because it is signed. So 0x80000000 >> 1 is 0xc0000000; >> 2 is 0xe0000000; >> 23 is 0xffffffff; >> 31 is 0xffffffff.

The value 10 is binary 1010, so it and to non-zero for every i in 28,29,30,31. You can test this by modifying your program to try some different numbers:

4: 00000000000000000000000000000111
8: 00000000000000000000000000000111
15:00000000000000000000000000001111
16:00000000000000000000000000011111

As the comments mention, if you switch to unsigned, this will not happen.

ps: It is entirely possible that signed shifting has been redefined to be undefined behaviour, in which case you are lucky your house didn’t burn down.