Decimal to Binary on C library - c

Question

I want to know if there is a function in C library that convert a decimal to binary number and save number by number in a position on an array.
For example: 2 -> 10 -> array [0] = 0 array[1] = 1.
Thanks.

Answer 1

here:
void dec2bin(int c)
{
int i = 0;
for(i = 31; i >= 0; i--){
if((c & (1 << i)) != 0){
printf("1");
}else{
printf("0");
}
}
}
But this only prints the value of an integer in binary format. All data is represented in binary format internally anyway.

Answer 2

You did not define what is a decimal number for you. I am guessing it is character representation (e.g. in ASCII) of that number.
Notice that numbers are just numbers. Binary or decimal numbers do not exist, but a given number may have a binary, and a decimal, representation. Numbers are not made of digits!
Then you probably want sscanf(3) or strtol(3) pr atoi to convert a string to an integer (e.g. an int or a long), and snprintf(3) to convert an integer to a string.
If you want to convert a number to a binary string (with only 0 or 1 char-s in it) you need to code that conversion by yourself. To convert a binary string to some long use strtol.

Answer 3

There is no such function in C standard library. Anyway, you can write your own:
void get_bin(int *dst, intmax_t x);
Where dst is the resulting array (with 1s and 0s), and x is the decimal number.
For example:
C89 version:
#include <limits.h>
void get_bin(int *dst, int x)
{
int i;
for (i = sizeof x * CHAR_BIT - 1; i >= 0; --i)
*dst++ = x >> i & 1;
}
C99 version:
/* C99 version */
#include <limits.h>
#include <stdint.h>
void get_bin(int *dst, intmax_t x)
{
for (intmax_t i = sizeof x * CHAR_BIT - 1; i >= 0; --i)
*dst++ = x >> i & 1;
}
It works as follow: we run through the binary representation of x, from left to right. The expression (sizeof x * CHAR_BIT - 1) give the number of bits of x - 1. Then, we get the value of each bit (*dst++ = x >> i & 1), and push it into the array.
Example of utilisation:
void get_bin(int *dst, int x)
{
int i;
for (i = sizeof x * CHAR_BIT - 1; i >= 0; --i)
*dst++ = x >> i & 1;
}
int main(void)
{
int buf[128]; /* binary number */
int n = 42; /* decimal number */
unsigned int i;
get_bin(buf, n);
for (i = 0; i < sizeof n * CHAR_BIT; ++i)
printf("%d", buf[i]);
return 0;
}

Answer 4

Here is a version that explicitly uses a string buffer:
#include <string.h>
const char *str2bin(int num, char buffer[], const int BUFLEN)
{
(void) memset(buffer, '\0', BUFLEN );
int i = BUFLEN - 1; /* Index into buffer, running backwards. */
int r = 0; /* Remainder. */
char *p = &buffer[i - 1]; /* buffer[i] holds string terminator '\0'. */
while (( i >= 0 ) && ( num > 0 )) {
r = num % 2;
num = num / 2;
*p = r + '0';
i--;
p--;
}
return (p+1);
}

Answer 5

Use char * itoa ( int value, char * str, int base );
Find more here ...

Answer 6

the function should go like this:
int dec2bin(int n){
static int bin,osn=1,c;
if(n==0) return 0;
else {
c=n%2;
bin += c*osn;
osn*=10;
dec2bin(n/2);
}
return bin;
}

Answer 7

As far as i know there is no such function in any C library. But here's a recursive function that returns a binary representation of a decimal number as int:
int dec2bin(int n)
{
if(n == 0) return 0;
return n % 2 + 10 * dec2bin(n / 2);
}
The max number that it can represent is 1023 (1111111111 in binary) because of int data type limit, but you can substitute int for long long data type to increase the range. Then, you can store the return value to array like this:
int array[100], i = 0;
int n = dec2bin(some_number);
do{
array[i] = n % 10;
n /= 10;
i++;
}while(n > 10)
I know this is an old post, but i hope this will still help somebody!

Answer 8

If it helps you can convert any decimal to binary using bitset library, for example:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main(){
int decimal = 20;
bitset<5> binary20(decimal);
cout << binary20 << endl;
return 0;
}
So, you have an output like 10100. Bitsets also have a "toString()" method for any purpose.