Divide a string of 32 into 4 8 hex in c

Question

community, I have little experience with C and I am on the learning curve right now.

I am working on a little project that involves dividing a 32-char string into 4 strings of 8 chars each in C. The 32-char string should resemble a 32-bit instruction. Those "32 bits" are divided into 4 "8-bit" strings that I want to print out as Hex. The code below is what I got so far. The data types I am using are the ones I am using in the rest of my code. I intend to feed the unsigned char t variable into an Substitution Box program that will give me equivalent of that t char from the S-Box lookup table.

The code below seems to me like it should work

unsigned char inst[] = "10101010101010101111111100111101";
unsigned char in[8];
for (int i = 0; i < 33; i++){
        if (i%8 == 0 && i != 0) {
                unsigned char t = (unsigned char) strtol(in, NULL, 2);
                printf("%x \n", t);
    }

    in[i%8] = inst[i];
    printf("%c ", in[i%8]);
}

but the output looks like this:

1 0 1 0 1 0 1 0 3d
1 0 1 0 1 0 1 0 3d
1 1 1 1 1 1 1 1 3d
0 0 1 1 1 1 0 1 3d

I can see that in[i%8] = inst[i]; line is reading the chars from inst[] correctly, but the

if (i%8 == 0 && i != 0) {
                unsigned char t = (unsigned char) strtol(in, NULL, 2);
                printf("%x \n", t);
    }

conditional statement prints the wrong hex. The output should look like something like this

1 0 1 0 1 0 1 0 aa
1 0 1 0 1 0 1 0 aa
1 1 1 1 1 1 1 1 ff
0 0 1 1 1 1 0 1 3d

Any help would be appreciated.

Answer 1

The problem is your in is not NUL terminated.

Thus passing in to strol invokes the undefined behavior.

Do as below.

    unsigned char in[9]; //+1 to hold the NUL char.

     ....
    if (i%8 == 0 && i != 0) {
                in[8] = '\0'; //NUL terminate the string.
                unsigned char t = (unsigned char) strtol(in, NULL, 2);
                printf("%x \n", t);
    }

Answer 2

Problems with the current code:

• "4 strings of 8 chars each" is char in[4][8+1]; and not char in[8]. You need room for null termination.
• 32 bits means iterate from 0 to 31, not from 0 to 32.
• There's no need to copy byte per byte. It's slow and makes everything needlessly complicated.

This seems to be the requirements:

• Split the original string in 4 sub strings.
• Convert each sub string to an integer.
• Display the result as hex

In which case you can simply iterate 4 times over the input string:

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

int main (void)
{
  const char* inst = "10101010101010101111111100111101";
  char in [4][8+1];

  puts("Bin      Hex");
  for(size_t i=0; i<4; i++)
  {
    memcpy(in[i], &inst[i*8], 8);
    in[i][8] = '\0';
    unsigned long val = strtoul(in[i], NULL, 2);    
    printf("%.8s %.2lX\n", in[i], val);
  }
}

Output:

Bin      Hex
10101010 AA
10101010 AA
11111111 FF
00111101 3D