conversion of uint8_t to a string [C]

Question

I'm trying to "translate" an array of uint8_t [uint8_t lets_try[16]] to a string of 16*8+1[null character] elements. For example:

lets_try[0] = 10101010  
lets_try[1] = 01010101  

...

and I would like to have a string like:

1010101001010101...[\0]

Here the questions: 1) is there a quick way to perform this operation?

I was trying to do it on my own; my idea was starting from translating a single uint8_t variable into a string and obtaining the full array with a loop [I haven't done this last part yet]. At the end I wrote this code:

int main()
{
    uint8_t example = 0x14;
    uint8_t *pointer;
    char *final_string;

    pointer = &example;

    final_string = convert(pointer);
    puts(final_string);

    return(0);
}


char *convert (uint8_t *a)
{
    int buffer1[9];
    char buffer2[9];
    int i;
    char *buffer_pointer;

    buffer1[8]='\0';

    for(i=0; i<=7; i++)
        buffer1[7-i]=( ((*a)>>i)&(0x01) );

    for(i=0; i<=7; i++)
        buffer2[i] = buffer1[i] + '0';

    buffer2[8] = '\0';

    puts(buffer2);

    buffer_pointer = buffer2;

    return buffer_pointer;
}

Here other few questions:

2) I'm not sure I fully understand the magic in this expression I found online: buffer2[i] = buffer1[i] + '0'; can somebody explain to me why the following puts(buffer2) is not going to work correctly without the +'0'? is it the null character at the end of the newborn string which makes the puts() work? [because with the null character it knows it's printing a real string?]

3) in the code above puts(buffer2) gives the right output while the puts in main() gives nothing; I'm going mad in looking again and again the code, I can't find what's wrong with that

4) in my solution I manage to convert an uint8_t into a string passing from an array of int: uint8_t->int array->string; is there a way to shorten this procedure, passing directly from the uint8_t into a string, or improve it? [in forums I found only solutions in C++] it works but I find it a little heavy and not so elegant

Thanks everybody for the support

Answer 1

1.) it's a little bit faster to eliminate the int array.

2.) adding '0' changes the integer values 0 and 1 to their ascii values '0' and '1'.

3.) it's undefined behaviour to return the address of a local variable. You have to malloc memory in the heap.

4.) yes, just cut it out and do the whole operation all in one

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

typedef unsigned char uint8_t;

char *convert(uint8_t *a)
{
  char* buffer2;
  int i;

  buffer2 = malloc(9);
  if (!buffer2)
    return NULL;

  buffer2[8] = 0;
  for (i = 0; i <= 7; i++)
    buffer2[7 - i] = (((*a) >> i) & (0x01)) + '0';

  puts(buffer2);

  return buffer2;
}


int main()
{
  uint8_t example = 0x14;
  char *final_string;

  final_string = convert(&example);
  if (final_string)
  {
    puts(final_string);

    free(final_string);
  }
  return 0;
}