CODEWITHSUNDEEP
cbit-manipulationbit
Christian Johnny Sotoparra
Christian Johnny Sotoparra

Reputation: 53

How do I change the bytes of a 32-bit integer?

I'm trying to add two binary numbers together using only logical statements and binary arithmetic operators. But I'm confused on how to actually change the bits. It is mostly the out variable that I am trying to change, but it keeps getting zeroed every time I print it. 

#include <stdio.h>

void execute_add(int a, int b){

int i = 0;
int bit;
int bit2;
int carryOut = 0;
int out = 10;
int overflow = 0;

for(i = 0; i <32 ; i++){

  bit = (a >> i) & 1;
  bit2 = (b >> i) & 1;

  if(bit==1 && bit2==1 && carryOut == 0){
        carryOut = 1;
        out = 0 | (0x1 >> i);

  }else if(bit==1 && bit2==1 && carryOut == 1){
        carryOut = 1;
        out = 1 | (0x1 >> i);

  }else if(bit==0 && bit2==0 && carryOut == 0){
        carryOut = 0;
        out= 0 | (0x1 >> i);

  }else if(bit==0 && bit2==0 && carryOut == 1){
        carryOut = 0;
        out = 1 | (0x1 >> i);

  }else if(bit==1 && bit2==0 && carryOut == 0){
        carryOut = 0;
        out = 1 | (0x1 >> i);

  }else if(bit==1 && bit2==0 && carryOut == 1){
        carryOut = 1;
        out = 0 | (0x1 >> i);

  }else if(bit==0 && bit2==1 && carryOut == 0){
        carryOut = 0;
        out = 1 | (0x1 >> i);

  }else if(bit==0 && bit2==1 && carryOut == 1){
        carryOut = 1;
        out = 0  | (0x1 >> i);

  }else{

  }//if else

 }//for loop

   printf("\n");

   bit = (a >> 31) & 1;
   bit2 = (a >> 31)& 1;
   int bit3 = (out >> 31) & 1;

   if( bit == 1 && bit2== 1 && bit3 == 0){
     overflow = 1;

   }else if (bit == 0 && bit2 == 0 && bit3 == 1){
   overflow = 1;

   }else{

   }//overflow check

    int j;
    int g = 0;

    for(j = 31; j>=0; j--){

            if(g%4==0 && g!=0){
             printf(" ");
            }

        bit2 = (out >> j) & 1;

        printf("%d", bit2);
        g++;
    }

            printf("\n");
}

int main (){

  int a = 34;
  int b = 17;

  execute_add(a, b);
  return 0;
}

Upvotes: 1

Views: 469

Answers (1)

Haldean Brown
Haldean Brown

Reputation: 12711

With each of these statements in your for loop:

out = 0 | x;

You're resetting out, and clearing away all the work you've already done. You probably mean to do:

out = out | x

Or, equivalently,

out |= x

You also are right-shifting 1 all over the place, which is not what you're looking for; for any shift greater than zero, that's going to give you zero. I think you're often looking for a left-shift where you use a right shift.

I would second kaylum's comment about using a debugger; even if you flip the appropriate shifts and | with out properly, you're still going to have logic errors that will be easily fixed with a debugger.

Upvotes: 1

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