Display the binary representation of a number in C?

Question

Still learning C and I was wondering:

Given a number, is it possible to do something like the following?

char a = 5;
printf("binary representation of a = %b",a);
> 101

Or would i have to write my own method to do the transformation to binary?

Answer 1

You have to write your own transformation. Only decimal, hex and octal numbers are supported with format specifiers.

Answer 2

There is no direct way (i.e. using printf or another standard library function) to print it. You will have to write your own function.

/* This code has an obvious bug and another non-obvious one :) */
void printbits(unsigned char v) {
   for (; v; v >>= 1) putchar('0' + (v & 1));
}

Answer 3

There is no direct format specifier for this in the C language. Although I wrote this quick python snippet to help you understand the process step by step to roll your own.

#!/usr/bin/python

dec = input("Enter a decimal number to convert: ")
base = 2
solution = ""

while dec >= base:
    solution = str(dec%base) + solution
    dec = dec/base
if dec > 0:
    solution = str(dec) + solution

print solution

Explained:

dec = input("Enter a decimal number to convert: ") - prompt the user for numerical input (there are multiple ways to do this in C via scanf for example)

base = 2 - specify our base is 2 (binary)

solution = "" - create an empty string in which we will concatenate our solution

while dec >= base: - while our number is bigger than the base entered

solution = str(dec%base) + solution - get the modulus of the number to the base, and add it to the beginning of our string (we must add numbers right to left using division and remainder method). the str() function converts the result of the operation to a string. You cannot concatenate integers with strings in python without a type conversion.

dec = dec/base - divide the decimal number by the base in preperation to take the next modulo

if dec > 0: solution = str(dec) + solution - if anything is left over, add it to the beginning (this will be 1, if anything)

print solution - print the final number

Answer 4

Yes (write your own), something like the following complete function.

#include <stdio.h> /* only needed for the printf() in main(). */
#include <string.h>

/* Create a string of binary digits based on the input value.
   Input:
       val:  value to convert.
       buff: buffer to write to must be >= sz+1 chars.
       sz:   size of buffer.
   Returns address of string or NULL if not enough space provided.
*/
static char *binrep (unsigned int val, char *buff, int sz) {
    char *pbuff = buff;

    /* Must be able to store one character at least. */
    if (sz < 1) return NULL;

    /* Special case for zero to ensure some output. */
    if (val == 0) {
        *pbuff++ = '0';
        *pbuff = '\0';
        return buff;
    }

    /* Work from the end of the buffer back. */
    pbuff += sz;
    *pbuff-- = '\0';

    /* For each bit (going backwards) store character. */
    while (val != 0) {
        if (sz-- == 0) return NULL;
        *pbuff-- = ((val & 1) == 1) ? '1' : '0';

        /* Get next bit. */
        val >>= 1;
    }
    return pbuff+1;
}

Add this main to the end of it to see it in operation:

#define SZ 32
int main(int argc, char *argv[]) {
    int i;
    int n;
    char buff[SZ+1];

    /* Process all arguments, outputting their binary. */
    for (i = 1; i < argc; i++) {
        n = atoi (argv[i]);
        printf("[%3d] %9d -> %s (from '%s')\n", i, n,
            binrep(n,buff,SZ), argv[i]);
    }

    return 0;
}

Run it with "progname 0 7 12 52 123" to get:

[  1]         0 -> 0 (from '0')
[  2]         7 -> 111 (from '7')
[  3]        12 -> 1100 (from '12')
[  4]        52 -> 110100 (from '52')
[  5]       123 -> 1111011 (from '123')

Answer 5

Use a lookup table, like:

char *table[16] = {"0000", "0001", .... "1111"};

then print each nibble like this

printf("%s%s", table[a / 0xF], table[a % 0xF]);

Surely you can use just one table, but it will be marginally faster and too big.

Answer 6

Based on dirkgently's answer, but fixing his two bugs, and always printing a fixed number of digits:

void printbits(unsigned char v) {
  int i; // for C89 compatability
  for(i = 7; i >= 0; i--) putchar('0' + ((v >> i) & 1));
}

Answer 7

Have a look here

Printing an unsigned integer in bits

Answer 8

#include<iostream>
#include<conio.h>
#include<stdlib.h>
using namespace std;
void displayBinary(int n)
{
       char bistr[1000];
       itoa(n,bistr,2);       //2 means binary u can convert n upto base 36
       printf("%s",bistr);

}

int main()
{
    int n;
    cin>>n;
    displayBinary(n);
    getch();
    return 0;
}