Thread: Decimal to binary conversion

Ive been trying for days to figure out how to do decimal to binary conversion... ive been trying to modules the number by 2 to get the remainder but my proble is saving the remainders in the right order to be printed... the asignment is as follows, print a table of the binary octal and hexadecimal equivalents from 1 to 256. ive got the octal and hexidecimal down... this is basic C programming.. im only in my 7th week of learning it, can someone please help?? here is the code i have already

Code:

#include <stdio.h>

int main()
{

  int x;

  for( x = 1; x <= 256; x++ ) {

    printf( "%d", x );
    printf( "\t" );
    printf( "%o", x );
    printf( "\t" );
    printf( "%x", x );
    printf( "\n" );
  }

 return 0;
}

Thanks for your help!!
Antoni

hope this bit helps in your case

Code:

#include<stdio.h>

int main()
{
        int num,i=0;
        int binary[10]={0};

        scanf("%d",&num);
        printf("Binary value of %d is ",num);
        while(num!=0)
        {
            
            binary[i++]=num%2;
            num/=2;
    }
    
    for(;i>=0;i--)
    printf("%d",binary[i]);
     
        getchar();
  }

this code calculates the binary value of the decimal value entered by the user

s.s.harish

You could do this too,

Code:

#include <stdio.h>
#define MASK 0x00000001

int main()
{
    int x, i;
   for(x =1; x <= 256; x++)
     {
        printf( "%d\t", x );
        printf( "%o\t", x );
        printf( "%x\t", x );

        for(i = 0; i < 33; i++)
          {
             printf("%i", (x && MASK));
             x = x >> 1;
          }
         printf("\n");
     }
  return 0;
}

Its bitwise operators which is not basic so if you do not want to use it thats ok.

Stumpster123: The way you print out the binary is still wrong since it goes in the wrong direction. So 2 is meant to be 0000 0010 but your code will print it as 0100 0000. A simple fix to this is to place the values into an array and print out the array in the reverse order once you have all the bits down.