使用按位运算符的模式

Question

How do I set the bits of a variable to assume the pattern I want? For exemple, if I have to print this sequence, how do i proceed?

11010010 11010010 11010010 11010010 

I wrote the code that print and separate bits in this configuration, but don't know how to set them as I want.

#include <stdio.h>
#include <limits.h>
int a;
void stampabit(a);
int main()
{
    int i;
    int n= sizeof(int) * CHAR_BIT;
    int mask = 1 << (n-1);

    for (i=1; i<=n; ++i){
        putchar(((a & mask)==0)?'0':'1');
        a<<=1;
        if(i%CHAR_BIT==0 && i<n)
            putchar(' ');
    }
}

Answer 1

You must shift mask instead of shift the variable

#include <stdio.h>
#include <limits.h>
unsigned int a = 0xAA55AA55;

int main()
{
    size_t i;
    unsigned int  n= sizeof(int) * CHAR_BIT;
    unsigned int  mask = 1 << (n-1);

    for (i=1; i<=n; ++i){
        putchar(((a & mask)==0)?'0':'1');
        mask>>=1;
        if(i%CHAR_BIT==0 && i<n)
            putchar(' ');
    }
    putchar('\n');
}

Output will be

10101010 01010101 10101010 01010101

Changing value of a to 0xD2D2D2D2 as you want output will be

11010010 11010010 11010010 11010010 

Answer 2

Edit:
if you need just set some bits, simply use bitwise or,
and if you need to clear some bits, use bitwise and:

uint32_t   a=1;
a |= 0xD2D2D2D2 ; //set bits to 11010010 11010010 11010010 11010010
a &= ~1; // clear first bit (or mask) 

and it is good to separate display code from main code to have clear code:

#include <stdio.h> 
#include <stdint.h>

void print_bits(uint32_t u){
    int i=31;
    do{
        if (u & 0x80000000) putchar('1'); else putchar('0');
        u <<= 1;
        if(i%8 == 0) putchar(' ');
    }while( --i >= 0);  
} 

int main(){  
    uint32_t   a=1;
    a |= 0xD2D2D2D2 ; //set bits to 11010010 11010010 11010010 11010010
    a &= ~1; // clear first bit (or mask) 
    print_bits(a); //output: 11010010 11010010 11010010 11010010
}

Old:
if you need just set some bits, simply use bitwise or like this:

void stampabit(int setBits){
    a |= setBits;
}

to set your pattern 11010010 11010010 11010010 11010010 use this:

stampabit(0xD2D2D2D2 ); // stampabit(0b11010010110100101101001011010010);

working example:

#include <stdio.h>
#include <limits.h>
//#include <stdint.h>
int a=0;
void stampabit(int setBits){
    a |= setBits;
}
int main()
{
    int i;
    stampabit(0xD2D2D2D2 ); // stampabit(0b11010010110100101101001011010010);
    int n= sizeof(int) * CHAR_BIT; // number of bits in int (8, 16, 32, 64, ... bits)
    int mask = 1 << (n-1); // sign bit mask

    for (i=1; i<=n; ++i){
        putchar(((a & mask)==0)?'0':'1');
        a<<=1;
        if(i%CHAR_BIT==0 && i<n)
            putchar(' ');
    }
}

output:

11010010 11010010 11010010 11010010

Some notes:
here it is just sample code and it works fine, but using unsigned types is clearly shows you don't need sign bit, and it is bug proof.
using local variables or shifting local variables is better than using or shifting global variables, unless it is intentional.

last but not least if you don't need platform dependent int use int32_t or uint32_t from

#include <stdint.h>

and if you need to clear some bits, use bitwise and:

uint32_t   a=1;
a |= 0xD2D2D2D2 ; //set bits to 11010010 11010010 11010010 11010010
a &= ~1; // clear first bit (or mask)

I hope this helps.