Converting C string to binary representation

Question

In ANSI C, how do we convert a string in to an array of binary bytes? All the googling and searching gives me answers for C++ and others and not C.

One idea I had was to convert the string into ASCII and then convert each ASCII value into its binary. (Duh!) I know it is the dumbest of ideas but I am not sure of any other option.

I've heard abt the encoding function in Java. I am not sure if that suits the same purpose and can be adopted to C.

string = "Hello"
bytearr[] = 10100101... some byte array..

It would be great if someone can throw some light on this.

Thanks!

What do you mean by "array of binary bytes" ? A "String" in C is simply a chunk of memory (an array) containing values (bytes) that get mapped to ASCII characters. — Brian Roach
– Brian Roach, Commented Apr 14, 2011 at 17:08
Some thing similar to the byte array in Java. Where u can process the string also in the form of a byte array. — Maverickgugu
– Maverickgugu, Commented Apr 14, 2011 at 17:10
You seem to be very confused about terminology. A string in C already is an array of binary bytes, more or less by definition. And it's probably also already ASCII (unless it's some other encoding of Unicode that supports characters outside U+0000 through U+007F). So please try again to explain what you want the contents of this "bytearr" to be. — zwol
– zwol, Commented Apr 14, 2011 at 17:11

Athabaska Dick · Accepted Answer · 2011-04-14 20:48:26Z

10

Or did you mean how to convert C string to binary representation?

Here is one solution which can convert strings to binary representation. It can be easily altered to save the binary strings into array of strings.

#include <stdio.h>

int main(int argc, char *argv[])
{
    if(argv[1] == NULL) return 0; /* no input string */

    char *ptr = argv[1];
    int i;

    for(; *ptr != 0; ++ptr)
    {
        printf("%c => ", *ptr);

        /* perform bitwise AND for every bit of the character */
        for(i = 7; i >= 0; --i) 
            (*ptr & 1 << i) ? putchar('1') : putchar('0');

        putchar('\n');
    }

    return 0;
}

Example input & output:

./ascii2bin hello

h => 01101000
e => 01100101
l => 01101100
l => 01101100
o => 01101111

edited Apr 14, 2011 at 20:48

answered Apr 14, 2011 at 17:43

Athabaska Dick

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Comments

Brian Roach · Accepted Answer · 2011-04-14 17:16:18Z

3

There is no any strings in C. Any string IS an array of bytes.

edited Apr 14, 2011 at 17:16

Brian Roach

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answered Apr 14, 2011 at 17:12

Jurlie

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1 Comment

Jurlie Over a year ago

I meant "Any string IS an array of bytes."

John Bode · Accepted Answer · 2011-04-14 17:15:28Z

1

A string is an array of bytes.

If you want to display the ASCII value of each character in hex form, you would simply do something like:

while (*str != 0)
  printf("%02x ", (unsigned char) *str++);

answered Apr 14, 2011 at 17:15

John Bode

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Comments

user257111 · Accepted Answer · 2011-04-14 17:15:38Z

1

On most of the systems I have worked on, the width of char is 1-byte and so a char[] or char* is a byte array.

In most other languages such as Java, the string datatype takes care of looking after, to a certain degree, concepts like encoding, by using an encoding like say UTF-8. In C this is not the case. If I were to read a UTF-8 string whose contents included multi-byte values, my characters would be represented by two buckets in the array (or potentially more).

To look at it from another point of view, consider that all types in C have a fixed width for your system (although they may vary between implementations).

So that string you're operating on is a byte array.

Next question I guess then is how do you display those bytes? That's pretty straightforward:

char* x = ???; /* some string */
unsigned int xlen = strlen(x);
int i = 0;

for ( i = 0; i < xlen; i++ )
{
    printf("%x", x[i]);
}

I can't think of a reason why you'd want to convert that output to binary, but it could be done if you were so minded.

answered Apr 14, 2011 at 17:15

user257111

1 Comment

zwol Over a year ago

This is not quite the same as "the width of char is 1 byte", but it probably deserves being said again in this context: sizeof(char)==1 BY DEFINITION. It will never be anything else. (However, the value of CHAR_BIT is not necessarily 8.)

zwol · Accepted Answer · 2011-04-14 17:13:58Z

0

If you just want to iterate (or randomly access) individual bytes' numeric values, you don't have to do any conversion at all, because C strings are arrays already:

void dumpbytevals(const char *str)
{
    while (*str)
    {
        printf("%02x ", (unsigned char)*str);
        str++;
    }
    putchar('\n');
}

If you're not careful with this kind of code, though, you run the risk of being in a world of hurt when you need to support non-ASCII characters.

answered Apr 14, 2011 at 17:13

zwol

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Comments

Nathan Moinvaziri · Accepted Answer · 2011-05-07 07:36:26Z

Since printf is slow when converting a huge binary array. Here is another approach that does not use printf:

#define BASE16VAL               ("x0x1x2x3x4x5x6x7x8x9|||||||xAxBxCxDxExF") 
#define BASE16_ENCODELO(b)      (BASE16SYM[((uint8)(b)) >> 4])
#define BASE16_ENCODEHI(b)      (BASE16SYM[((uint8)(b)) & 0xF]) 
#define BASE16_DECODELO(b)      (BASE16VAL[Char_Upper(b) - '0'] << 4)
#define BASE16_DECODEHI(b)      (BASE16VAL[Char_Upper(b) - '0']).

To convert a hex string to a byte array you would do the following:

while (*Source != 0)   
    {   
    Target[0]  = BASE16_DECODELO(Souce[0]);   
    Target[0] |= BASE16_DECODEHI(Souce[1]);    

    Target += 1;   
    Source += 2;   
    } 

*Target = 0;

Source is a pointer to a char array that contains a hex string. Target is a pointer to a char array that will contain the byte array.

To convert a byte array to a hex string you would to the following:

while (*Source != 0)   
    {   
    Target[0] = BASE16_ENCODELO(*Source);   
    Target[1] = BASE16_ENCODEHI(*Source);    

    Target += 2;   
    Source += 1;   
    }

Target is a pointer to a char array that contains a hex string. Source is a pointer to a char array that will contain the byte array.

Here are a few missing macros:

#define Char_IsLower(C)  ((uint8)(C - 'a') < 26)
#define Char_IsUpper(C)  ((uint8)(C - 'A') < 26)
#define Char_Upper(C)    (Char_IsLower(C) ? (C + ('A' - 'a')) : C)
#define Char_Lower(C)    (Char_IsUpper(C) ? (C + ('a' - 'A')) : C)

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