I have the following code:

pid_t pid = fork();
if (pid == -1)
{
    // ...
}
else if (pid == 0)
{
    stdin = someopenfile;
    stdout = someotherfile;
    stderr = somethirdopenfile;
    execvp(args[0], args);
    // handle error ...
}
else
{
    // ...
}

The problem is, the input/output of the execvp() call is still the console, rather than the files. Clearly I am doing something wrong, what is the right way to do this?

Solution 1

The right way to do it is to replace the file descriptors STDIN_FILENO, STDOUT_FILENO and STDERR_FILENO with the opened files using dup2(). You should also then close the original files in the child process:

else if (pid == 0)
{
    dup2(fileno(someopenfile), STDIN_FILENO);
    dup2(fileno(someotherfile), STDOUT_FILENO);
    dup2(fileno(somethirdopenfile), STDERR_FILENO);
    fclose(someopenfile);
    fclose(someotheropenfile);
    fclose(somethirdopenfile);
    execvp(args[0], args);
    // handle error ...
}

Solution 2

Take a look at freopen function.

I had to do something similar with stdout and wrote two functions that do the work for me:

static int fd;
static fpos_t pos;

void switchStdout(const char *newStream)
{
  fflush(stdout);
  fgetpos(stdout, &pos);
  fd = dup(fileno(stdout));
  freopen(newStream, "w", stdout);
}

void revertStdout()
{
  fflush(stdout);
  dup2(fd, fileno(stdout));
  close(fd);
  clearerr(stdout);
  fsetpos(stdout, &pos);
}

Solution 3

You can use this when stdin , stdout , stderr are terminal-

//change stdin,stdout,stderr
    freopen("new_stdin","r",stdin);
    freopen("new_stdout","r",stdout);
    freopen("new_stderr","r",stderr);

    //----do something;

//reset stdin,stdout,stderr
     freopen("/dev/tty","r",stdin);
     freopen("/dev/tty","r",stdout);
     freopen("/dev/tty","r",stderr);