Converting massive binary input string into character string C

Question

I'm not familiar with C at all so this might be a simple problem to solve. I'm trying to take an input char* array of binary character sequences, ex. "0100100001101001", and output its relative string ("Hi"). The problem I'm having is coming up with a way to split the input into seperate strings of length 8 and then convert them individually to ulimately get the full output string.

char* binaryToString(char* b){
    char binary[8];
    for(int i=0; i<8; ++i){
    binary[i] = b[i];
}
printf("%s", binary);
}

I'm aware of how to convert 8-bit into its character, I just need a way to split the input string in a way that will allow me to convert massive inputs of 8-bit binary characters.

Any help is appreciated... thanks!

1001001011101000010011101110011001000000111011101101111011100100110101101101001011011100110011100100001 — Steve Summit, Dec 05 '17 at 04:13

score 2 · Accepted Answer · edited Dec 07 '17 at 01:31

From what I can tell, your binaryToString() function does not do what you'd want it to. The print statement just prints the first eight characters from the address pointed to by char* b.

Instead, you can convert the string of 8 bits to an integer, utilizing a standard C function strtol(). There's no need to convert any further, because binary, hex, decimal, etc, are all just representations of the same data! So once the string is converted to a long, you can use that value to represent an ASCII character.

Updating the implementation (as below), you can then leverage it to print a whole sequence.

#include <stdio.h>
#include <string.h>

void binaryToString(char* input, char* output){

    char binary[9] = {0}; // initialize string to 0's

    // copy 8 bits from input string
    for (int i = 0; i < 8; i ++){
        binary[i] = input[i];    
    }

    *output  = strtol(binary,NULL,2); // convert the byte to a long, using base 2 
}

int main()
{

    char inputStr[] = "01100001011100110110010001100110"; // "asdf" in ascii 
    char outputStr[20] = {0}; // initialize string to 0's

    size_t iterations = strlen(inputStr) / 8; // get the # of bytes

    // convert each byte into an ascii value
    for (int i = 0; i < iterations; i++){
        binaryToString(&inputStr[i*8], &outputStr[i]);
    }

    printf("%s", outputStr); // print the resulting string
    return 0;
}

I compiled this and it seems to work fine. Of course, this can be done cleaner and safer, but this should help you get started.

Yes I just set up the print for testing purposes... I must've accidentally took the function from a wrong file. — T. Pruden, Dec 05 '17 at 02:27

burakozgul · Answer 2 · 2022-12-12T08:40:28.510

Maybe this can help. I didnt compile it but there is the idea. You can loop every 8 bit separately with while loop. And assign 8 bit to binary array with for loop. After that send this binary array to convert8BitToChar function to get letter equivalent of 8 bit. Then append the letter to result array. I'm not writing c for 3 year if there is mistakes sorry about that. Here pseudo code.

char* binaryToString(char* b){
char* result = malloc(sizeof(256*char));
char binary[8];
int nextLetter = 0;
while (b[nextLetter*8] != NULL) { // loop every 8 bit 
    for(int i=0; i<8; ++i){
        binary[i] = b[nextLetter*8+i];
    }
    result[nextLetter] = 8bitToChar(binary));// convert 8bitToChar and append yo result
    nextLetter++;
}
result[nextLetter] = '\0';
return result;
}

`sizeof(256*char)` is same as `sizeof(int)` - insufficient for this task. Suggest `result = malloc(sizeof *result * 256);` instead. — chux - Reinstate Monica, Dec 05 '17 at 04:26

H.S. · Answer 3 · 2017-12-05T02:40:06.337

I just need a way to split the input string in a way that will allow me to convert massive inputs of 8-bit binary characters.

You can use strncpy() to copy the sequence of '0' and '1' in a chunk of 8 characters at a time from the input string, something like this:

//get the size of input string
size_t len = strlen(b);

//Your input array of '0' and '1' and every sequence of 8 bytes represents a character
unsigned int num_chars = len/8;

//Take a temporary pointer and point it to input string
const char *tmp = b;

//Now copy the 8 chars from input string to buffer "binary"
for(int i=0; i<num_chars; ++i){
    strncpy(binary, tmp+(i*8), 8);

    //do your stuff with the 8 chars copied from input string to "binary" buffer

}

Converting massive binary input string into character string C

3 Answers3