Thread: binary to decimal

  1. #1
    Registered User
    Join Date
    Mar 2008
    Posts
    33

    binary to decimal

    Hey,

    I can change decimal numbers to binary but I am having trouble doing it other way around...

    Code:
    int main()
    {
        int number, remainder;
        int highBitValue, numBits, bit;
        int i;
            
        printf("Please enter a decimal number: ");
        scanf("%d", &number);
        fflush(stdin);
         
       
      
        printf("The equivalent representation of %d in binary is ", number);  
                
        /* find the largest power of 2 that fits into number */
        highBitValue = 1; numBits = 0;
        remainder = number;
        
        while (remainder != 0)
        {
              remainder = remainder / 2;
              
              /* we keep track of both the highest digit value and the number of digits */
              highBitValue = highBitValue * 2;
              numBits++;
        } 
      
        
        remainder = number;
        
        /* avoid leading 0 bit */
        highBitValue = highBitValue / 2; 
             
        for (i=0; i<numBits; i++)
        {
              /* find the highest bit in remainder */
              bit = remainder / highBitValue;
              printf("%d", bit);
                
              /* get ready for the next bit */
              remainder = remainder - bit * highBitValue;    
              highBitValue = highBitValue / 2; 
        }
    Thank you.

  2. #2
    Hurry Slowly vart's Avatar
    Join Date
    Oct 2006
    Location
    Rishon LeZion, Israel
    Posts
    6,794
    fflush(stdin) is undefined - see FAQ

    bit manipulations better to do using bit-oparations like | ~ & >> <<

    show your attempt to convert binary to decimal and we will help to fix errors...

    or search form for samples - there are a lot (twice per month at least this question is discussed)
    All problems in computer science can be solved by another level of indirection,
    except for the problem of too many layers of indirection.
    David J. Wheeler

  3. #3
    Registered User
    Join Date
    Mar 2008
    Posts
    33
    hi, I finished writing the code but it seems I am trying to hard.. I am trying it to make it less complicated but this is what I got so far,

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    
    int main()
    {
        int n,i,temp1,temp2,temp3,temp4,temp5,remainder,j,k,p,count,count2,a;
        double bits;
    	bits = pow(2,n);
    
    	printf("Please enter an integer between 1 and 33:");
    	scanf("%d",&n);
    
        bits = pow(2,n);
     
    	printf("\nNumber n of bits in the integers: %d\n",n);
    	printf("Number of integers with n=4 bits: %d\n",bits);
    
    	printf("\n\nAll integers with 0 bits equal to 1 listed in ascending order:\n");
    
    	i = 1;	
        temp1 = n;
        temp2 = n;
        
        
         printf("0\t");
        
    	while (temp1 != 0)
    	{
              printf("0");
    		  temp1--;
    	}
    
    	printf("\n\n");
    
    	int l;
    	
        for(l=1;l<=n;l++)
    	{
             printf("All integers with %d bits equal to 1 listed in ascending order:\n",l);
             i=1;
             
             while(i != bits)
             {
                     
                     int temp3 = i;
                     int temp4 = i;
                     int count = 0;
                     
                     while(temp3 != 0)
                     {
                                    temp3 = temp3 / 2;
                                    count++;
                                    
                     }
                     
                     int k = count - 1;
                     
                     int sum = 0;
                     int count3 = 0;
                     
                     while (count != 0)
                     {
                           int power1 = (int) pow(2,k);
                           double power2 = pow(10, k);
                           
                           int a = temp4 / power1;
                           
                           temp4 = temp4 % power1;
                           sum = sum + a * power2;
                           
                           if(a == 1)
                           {
                                count3++;
                           }
                           
                           k--;
                           count --;
                     }     
                           int need1 = sum;
                           
                           
                           if (count3 == l)
                           {
                                      int f;
                                      
                                      printf("%d\t",i);
                                      
                                      count2=0;
                           
                                      while (need1 != 0)
                                      {
                                            need1 = need1 / 10;
                                            count2++;
                                             
                                      }
                                      f = n - count2;
                                      
                                      
                                      while(f != 0)
                                      {
                                              printf("0");
                                              f--;
                                              
                                      }   
                                      printf("%d\n",sum);
                           
                           }
                           
                     
                     i++;
             }
        printf("\n");    
        }
        system("pause");
    }
    can you help me make this more understandable for other people to understand without having to trace its every move =S
    Last edited by yigster; 03-31-2009 at 04:04 AM.

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