How to convert ascii string to binary?

Question

I'm trying to convert an ascii string to a binary string in C. I found this example Converting Ascii to binary in C but I rather not use a recursive function. I tried to write an iterative function as opposed to a recursive function, but the binary string is missing the leading digit. I'm using itoa to convert the string, however itoa is a non standard function so I used the implementation from What is the proper way of implementing a good "itoa()" function? , the one provided by Minh Nguyen.

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>

int32_t ascii_to_binary(char *input, char **out, uint64_t len)
{
    uint32_t i;
    uint32_t str_len = len * 8;

    if(len == 0)
    {
        printf("Length argument is zero\n");
        return (-1);
    }

    (*out) = malloc(str_len + 1);
    if((*out) == NULL)
    {
        printf("Can't allocate binary string: %s\n", strerror(errno));
        return (-1);
    }

    if(memset((*out), 0, (str_len)) == NULL)
    {
        printf("Can't initialize memory to zero: %s\n", strerror(errno));
        return (-1);
    }

    for(i = 0; i < len; i++)
        itoa((int32_t)input[i], &(*out)[(i * 8)], 2);

    (*out)[str_len] = '\0';

    return (str_len);
}

int main(void)
{
    int32_t rtrn = 0;
    char *buffer = NULL;

    rtrn = ascii_to_binary("a", &buffer, 1);
    if(rtrn < 0)
    {
        printf("Can't convert string\n");
        return (-1);
    }

    printf("str: %s\n", buffer);

    return (0);
}

I get 1100001 for ascii character a, but I should get 01100001, so how do I convert the ascii string to the whole binary string?

Just a comment: you don't need to test the result of `memset`. — Ian Abbott, Jan 18 '16 at 18:17
`itoa` does not procude leading zeroes. For example, with base 10 you will never get `097`. — Jongware, Jan 18 '16 at 18:19
@Jongware so `itoa` won't put a leading zero but it will put a leading one? — 2trill2spill, Jan 18 '16 at 18:23
Uh yahh. There is a big difference, isn't it? Any number can have a zillion leading zeroes and it won't change a thing. But add or leave out a one (or 4, or `F` when working in hex) and *it's value changes*. — Jongware, Jan 18 '16 at 18:26
@Jongware Yea there's a huge difference, I was just trying to make sure. — 2trill2spill, Jan 18 '16 at 18:28
Curious, why pass in length type `uint64_t len` and then form a narrower `uint32_t str_len`? suspect code should be using `size_t` for those two. — chux - Reinstate Monica, Jan 18 '16 at 18:52
1) Best not to "fix" this post though - use in the future. 2) I was hoping to find out why you used those types — chux - Reinstate Monica, Jan 18 '16 at 19:34
@chux It's just habit for me to have functions return either `int32_t` or a pointer type, that way the only values used for returning errors is `-1` or `NULL`. I will just make a fourth arg for the new size for later use. — 2trill2spill, Jan 18 '16 at 19:45

Ian Abbott · Accepted Answer · 2016-01-18T18:35:13.397

2

You could change the for loop to something like this:

for(i = 0; i < len; i++) {
    unsigned char ch = input[i];
    char *o = *out + 8 * i;
    int b;

    for (b = 7; b >= 0; b--)
        *o++ = (ch & (1 << b)) ? '1' : '0';
}

or similar:

for(i = 0; i < len; i++) {
    unsigned char ch = input[i];
    char *o = &(*out)[8 * i];
    unsigned char b;

    for (b = 0x80; b; b >>= 1)
        *o++ = ch & b ? '1' : '0';
}

edited Jan 18 '16 at 18:35

answered Jan 18 '16 at 18:26

Ian Abbott

15,083
19
33

neither of these crash with Address sanitizer on, it was one of your edits that was crashing. Both of these examples work fine, thanks. – 2trill2spill Jan 18 '16 at 18:52

score 0 · Answer 2 · answered Jan 18 '16 at 18:21

This program gets and integer ( which contains 32 bits ) and converts it to binary, Work on it to get it work for ascii strings :

#include <stdio.h>

int main()
{
  int n, c, k;

  printf("Enter an integer in decimal number system\n");
  scanf("%d", &n);

  printf("%d in binary number system is:\n", n);

  for (c = 31; c >= 0; c--)
  {
    k = n >> c;

    if (k & 1)
      printf("1");
    else
      printf("0");
  }

  printf("\n");

  return 0;
}

Nunchy · Answer 3 · 2016-01-18T18:40:43.893

-1

Best just write a simple function to do this using bitwise operators...

#define ON_BIT = 0x01

char *strToBin(char c) {
    static  char    strOutput[10];
    int     bit;

    /*Shifting bits to the right, but don't want the output to be in reverse
     * so indexing bytes with this...
     */
    int     byte;

    /* Add a nul at byte 9 to terminate. */
    strOutput[8] = '\0';

    for (bit = 0, byte = 7; bit < 8; bit++, byte--) {
        /* Shifting the bits in c to the right, each time and'ing it with
         * 0x01 (00000001).
         */
        if ((c >> bit) & BIT_ON)
            /* We know this is a 1. */
            strOutput[byte] = '1';
        else
            strOutput[byte] = '0';
    }

    return strOutput;
}

Something like that should work, there's loads of ways you can do it. Hope this helps.

edited Jan 18 '16 at 18:40

answered Jan 18 '16 at 18:23

Nunchy

948
5
11

Why are you returning `char strOutput[10];` from a function that returns just a char? – 2trill2spill Jan 18 '16 at 18:36
Heh, good point - that should return a char * - I stand corrected, cheers. – Nunchy Jan 18 '16 at 18:39
The edited function looks fine now, although it is not reentrant, and old results are overwritten every time it is called, unless the caller copies them elsewhere. – Ian Abbott Jun 08 '16 at 11:51

How to convert ascii string to binary?

3 Answers3