Convert decimal to binary in C

Question

I am trying to convert a decimal to binary such as 192 to 11000000. I just need some simple code to do this but the code I have so far doesn't work:

void dectobin(int value, char* output)
{
    int i;
    output[5] = '\0';
    for (i = 4; i >= 0; --i, value >>= 1)
    {
        output[i] = (value & 1) + '0';
    }
}

Any help would be much appreciated!

Answer 1

The value is not decimal. All values in computer's memory are binary.

What you are trying to do is to convert int to a string using specific base. There's a function for that, it's called itoa. http://www.cplusplus.com/reference/cstdlib/itoa/

Answer 2

First of all 192cannot be represented in 4 bits

192 = 1100 0000 which required minimum 8 bits.

Here is a simple C program to convert Decimal number system to Binary number system

#include <stdio.h>  
#include <string.h>  

int main()  
{  
    long decimal, tempDecimal;  
    char binary[65];  
    int index = 0;  

    /* 
     * Reads decimal number from user 
     */  
    printf("Enter any decimal value : ");  
    scanf("%ld", &decimal);  

    /* Copies decimal value to temp variable */  
    tempDecimal = decimal;  

    while(tempDecimal!=0)  
    {  
        /* Finds decimal%2 and adds to the binary value */  
        binary[index] = (tempDecimal % 2) + '0';  

        tempDecimal /= 2;  
        index++;  
    }  
    binary[index] = '\0';  

    /* Reverse the binary value found */  
    strrev(binary);  

    printf("\nDecimal value = %ld\n", decimal);  
    printf("Binary value of decimal = %s", binary);  

    return 0;  
} 

Answer 3

5 digits are not enough for your example (192). Probably you should increase output

Answer 4

A few days ago, I was searching for fast and portable way of doing sprintf("%d", num). Found this implementation at the page itoa with GCC:

/**
 * C++ version 0.4 char* style "itoa":
 * Written by Lukás Chmela
 * Released under GPLv3.

 */
char* itoa(int value, char* result, int base) {
    // check that the base if valid
    if (base < 2 || base > 36) { *result = '\0'; return result; }

    char* ptr = result, *ptr1 = result, tmp_char;
    int tmp_value;

    do {
        tmp_value = value;
        value /= base;
        *ptr++ = "zyxwvutsrqponmlkjihgfedcba9876543210123456789abcdefghijklmnopqrstuvwxyz" [35 + (tmp_value - value * base)];
    } while ( value );

    // Apply negative sign
    if (tmp_value < 0) *ptr++ = '-';
    *ptr-- = '\0';
    while(ptr1 < ptr) {
        tmp_char = *ptr;
        *ptr--= *ptr1;
        *ptr1++ = tmp_char;
    }
    return result;
}

Answer 5

It looks like this, but be careful, you have to reverse the resulting string :-)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char output[256]="";

int main()
{
int x= 192;
int n;
n = x;
int r;
do {
r = n % 2;
if (r == 1)
   strcat(output,"1");
else strcat(output,"0");
n = n / 2;
}
while (n > 0);

printf("%s\n",output);
}

Answer 6

So... did you check the output of your code to understand why it doesn't work?

So iteration 1 of your loop:
value = 192
i = 4
output[i] = (11000000 & 1) + '0' = 0 + 48 = 48 (char `0`)

Iteration 2 of your loop:
value = 96
i = 3
output[i] = (1100000 & 1) + '0' = 0 + 48 = 48 (char `0`)

Iteration 3 of your loop:
value = 48
i = 2
output[i] = (110000 & 1) + '0' = 0 + 48 = 48 (char `0`)

Iteration 4 of your loop:
value = 24
i = 1
output[i] = (11000 & 1) + '0' = 0 + 48 = 48 (char `0`)

Iteration 5 of your loop:
value = 12
i = 0
output[i] = (1100 & 1) + '0' = 0 + 48 = 48 (char `0`)

Final string: "00000"  and you wanted: "11000000"

See anything wrong with your code? Nope. Neither do I you just didn't go far enough. Change your output/loop to:

output[8] = '\0';
for (i = 7; i >= 0; --i, value >>= 1)

And then you'll have the correct result returned.

I would recomend just a more general approach, you're using a fixed length string, which limits you to binary numbers of a certian length. You might want to do something like:

loop while number dividing down is > 0
count number of times we loop
malloc an array the correct length and be returned

Answer 7

#include <stdio.h>
#include <stdlib.h>
void bin(int num) {
    int n = num;
    char *s = malloc(sizeof(int) * 8);
    int i, c = 0;
    printf("%d\n", num);

    for (i = sizeof(int) * 8 - 1; i >= 0; i--) {
        n = num >> i;
        *(s + c) = (n & 1) ? '1' : '0';
        c++;
    }
    *(s + c) = NULL;
    printf("%s", s); // or you can also return the string s and then free it whenever needed
}

int main(int argc, char *argv[]) {
    bin(atoi(argv[1]));
    return EXIT_SUCCESS;
}

Answer 8

You can do it using while loop under a function also. I was just searching the solve for mine but the solves i get were not suitable, so I have done it accordingly the practical approach (divide using 2 until getting 0 and store the reminder in an array) and print the reverse of the array and Shared Here

#include <stdio.h>

    int main()
    {
        long long int a,c;
        int i=0,count=0;
        char bol[10000];
        scanf("%lld", &a);
        c = a;
        while(a!=0)
        {
            bol[i] = a%2;
            a = a / 2;
            count++;
            i++;
        }
        if(c==0)
        {
            printf("0");
        }
        else
        {
            for(i=count-1; i>=0; i--)
            {
                printf("%d", bol[i]);
            }
        }
        printf("\n");
        return 0;
    }

Answer 9

Convert Decimal to Binary in C Language

#include<stdio.h>
void main()
{
    long int n,n1,m=1,rem,ans=0;
    printf("\nEnter Your Decimal No (between 0 to 1023) :: ");
    scanf("%ld",&n);

    n1=n;
    while(n>0)
    {
        rem=n%2;
        ans=(rem*m)+ans;
        n=n/2;
        m=m*10;
    }

    printf("\nYour Decimal No is   :: %ld",n1);
    printf("\nConvert into Binary No is :: %ld",ans);
}

Answer 10

//C Program to convert Decimal to binary using Stack

#include<stdio.h>

#define max 100

int stack[max],top=-1,i,x;  


void push (int x)
{
  ++top;
  stack [top] = x;
}

int pop ()
{ 
   return stack[top];
}   


void  main()
{
  int num, total = 0,item;
  print f( "Please enter a decimal: ");
  scanf("%d",&num);
  while(num > 0)
  {  
    total = num % 2;
    push(total);
    num /= 2;
  }



 for(i=top;top>-1;top--)
 {     
     item = pop ();
     print f("%d",item);
 }

 }

Answer 11

This is the simplest way to do it

#include <stdio.h>

void main()
{
    int n,i,j,sum=0;
    printf("Enter a Decimal number to convert it to binary : ");
    scanf("%d",&n);
    for(i=n,j=1;i>=1;j*=10,i/=2)
        sum+=(i%2)*j;
    printf("\n%d",sum);
}

Answer 12

This is a simple program to convert a number from decimal to binary

#include <stdio.h>
#include <conio.h>
void decToBinary(int);

int main()
{
    int number;

    printf("Enter number to convert to binary: ");
    scanf("%d", &number);
    
    decToBinary(number);
  return 0;
}

void decToBinary(int num)
{
    if (num == 0)
    {
        return ;
    }

    decToBinary(num / 2);
    printf("%d", num % 2);
}

Output of the Program:

Answer 13

Perhaps understanding the algorithm would allow you write or modify your own code to suit what you need. I do see that you don't have enough char array length to display your binary value for 192 though (You need 8 digits of binary, but your code only gives 5 binary digits)

Here's a page that clearly explains the algorithm.

I'm not a C/C++ programmer so here's my C# code contribution based on the algorithm example.

int I = 0;
int Q = 95;
string B = "";
while (Q != 0)
{
    Debug.Print(I.ToString());
    B += (Q%2);
    Q = Q/2;
    Debug.Print(Q.ToString());
    I++;
}

Debug.Print(B);

All the Debug.Print is just to show the output.

Answer 14

//decimal to binary converter
long int dec2bin(unsigned int decimal_number){
  if (decimal_number == 0)
    return 0;
  else
    return ((decimal_number%2) + 10 * dec2bin(decimal_number/2));
}

Answer 15

number=215
a=str(int(number//128>=1))+str(int(number%128>=64))+
str(int(((number%128)%64)>=32))+str(int((((number%12
8)%64)%32)>=16))+str(int(((((number%128)%64)%32)%16)>=8))
+str(int(((((((number%128)%64)%32)%16)%8)>=4)))
+str(int(((((((((number%128)%64)%32)%16)%8)%4)>=2))))
+str(int(((((((((((number%128)%64)%32)%16)%8)%4)%2)>=1)))))
print(a)

You can also use the 'if', 'else', statements to write this code.

Answer 16

int main() 
{ 
 int n, c, k;
 printf("Enter an integer in decimal number system: ");
 scanf("%d", &n);
 printf("%d in binary number system is: ", n);
 for (c = n; c > 0; c = c/2) 
  {
   k = c % 2;//To
   k = (k > 0) ? printf("1") : printf("0");
  }
 getch();
 return 0; 
}