LinuxQuestions.org - Opening a File by using a named pipe

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Opening a File by using a named pipe

Hi,

I am trying to use a function from ffmpeg library(A linux based decoder-encoder)

This function has an input like "const char *filename"

I am trying to call it with a buffer(unsigned char *buff)
So, I created a pipe, and wrote my buffer to it:

Code:

int my_pipe[2];

if(pipe(my_pipe) == -1)

{    perror("pipe call error");

        return -1;

}



write(my_pipe[1],buff,buff_size);

now, is there a way that instead of "filename", I pass my pipe to this function to read from?

(An equivalent question can be, how to call `fopen` using a file descriptor?)

I dont know if this is portable to other OSs, but it works in Linux.

Each process has its own directory, placed in /proc and named after its process ID (thus all the directories with numbers). Each process-dir has a directory named 'fd', which contains a list of all active file descriptors. /proc/self is a shortcut to the current process-dir (so it wont work to pass such a name between multiple processes, you would have to construct the actual name with the process ID).

Simple make a filename using the above and the file descriptor, example:

Code:

#include <stdio.h>

#include <stdlib.h>

#include <string.h>



int main (argc, argv)

  int argc;

  char** argv;

{

    char* buf;

    buf = "apple\n";



    int my_pipe[2];

    if(pipe(my_pipe) == -1)

      {

        perror("pipe call error");

        return -1;

      }



    write(my_pipe[1], buf, strlen(buf));



    FILE* file;

    char* filename;

    asprintf (&filename, "/proc/self/fd/%i", my_pipe[1]);

    file = fopen (filename, "r");

    if (file == NULL)

      {

        perror ("unable to open file");

        return -1;

      }

    free (filename);



    char* readbuf;

    int readsize;

    readbuf = NULL;

    readsize = 0;



    getline (&readbuf, &readsize, file);

    printf ("%s", readbuf);

    free (readbuf);



    fclose (file);

    close (my_pipe[0]);



    return 0;

}