Convert buffer to array of 8-bit values and back in C - c

Question

I would like to write a program in C that gets the file content via stdin and reads it line by line and, for each line, converts it to an array of 8-bit integer values.
I also would like to be able to do the reverse process. After working with my array of 8-bit values, I would like to convert it again to "lines" that would be organized as a new buffer.
So basically, I would like to convert a char * line to an int array[] and back (an int array[] to a char * line) while keeping the consistency, so when I create the file again out of the conversions, the file is valid (and by valid I mean, the conversion from int array[] to char * line generates the same content of the original char * line, while reading each line of the stdin.
My code is currently as follows:
#include <stdio.h>
#include <stdlib.h>
int main() {
FILE *stream;
char *line = NULL;
size_t len = 0;
ssize_t read;
stream = stdin;
if (stream == NULL)
exit(EXIT_FAILURE);
while ((read = getline(&line, &len, stream)) != -1) {
char * array = line_to_array(line);
// here I include the rest of my code
// where I am going to use the generated array
// ...
}
free(line);
fclose(stream);
exit(EXIT_SUCCESS);
}
The line_to_array function would be the one to convert the "line" content to the array of integers. In a second file, I would just do the opposite.
The mechanics of the process would be like this:
The first program (first.c) would receive a file content via stdin. By reading it using getline, I would have each line to convert to an array of integers and send each line to a second program (second.c) that would convert each array to a char * buffer again and the reconstruct the file.
In the terminal, I would run it like this:
./first | ./second
I appreciate any help on this matter.
Thank you.

Answer 1

I believe you may already know that a name of array is a kind of constant pointer. You could verify the fact from following code:
char hello[] = "hello world!";
for( int idx=0; *(hello + idx) != 0; idx++ )
{
printf("%c", *(hello + idx));
}
printf("\n");
So, there are no reason to convert character pointer to array. For your information, A char variable is a 8bit data in C, this can contain a integer value which is represent a character: 65 represent 'A' in ASCII code.
Secondly, this link may help you to understand how to convert between c string and std::string.
On second thought, may your input file is UNICODE or UTF-8 encoded file which is using multi-byte character code. In that case, you may not able to use getline() to read the string from the file. If so, please refer this question: Reading unicode characters.
I wish following code assist you to understand char type, array and pointer in C/C++:
std::string hello("Hello world");
const char *ptr = hello.c_str();
for( int idx=0; idx < hello.size(); idx++ )
{
printf("%3d ", *(ptr + idx));
}
printf("\n");
std::string hello("Hello world");
const char *ptr = hello.c_str();
for( int idx=0; idx < hello.size(); idx++ )
{
printf("%3d ", ptr[idx]);
}
printf("\n");