Convert decimal to binary in C - 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

//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 10

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 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;
}