Question

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Is there a printf converter to print in binary format?

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

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')