I am trying to catch the output from running the command /bin/lsx -lah /.

The output should be: bash: /bin/lsx: no such file or directory

However, I am just not getting anything in the printf().

This is the code

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

#define die(e) do { fprintf(stderr, "%s\n", e); exit(EXIT_FAILURE); } while (0);

int main() {
  int link[2];
  pid_t pid;
  char foo[4096];

  if (pipe(link)==-1)
    die("pipe");

  if ((pid = fork()) == -1)
    die("fork");

  if(pid == 0) {

    dup2 (link[1], STDOUT_FILENO);
    close(link[0]);
    close(link[1]);
    execl("/bin/lsx", "ls", "-1", (char *)0);
    die("execl");

  } else {

    close(link[1]);
    int nbytes = read(link[0], foo, sizeof(foo));
    printf("Output: (%.*s)\n", nbytes, foo);
    wait(NULL);

  }
  return 0;
}

I am just wondering why the output is not being captured and printed in the bottom printf().

CodePudding user response：

As John pointed out, you are only capturing stdout, not stderr, and well-formed programs usually send error messages to stderr (the die macro you have shown is an example of this).

A quick solution is to redirect the child processes' stderr to its stdout with another call to dup2

dup2(link[1], STDOUT_FILENO);
dup2(STDOUT_FILENO, STDERR_FILENO);

which will redirect both output streams to the single pipe.

Alternatively, capture both streams with two pipes to retain their separation as they propagate.

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>

#define die(e) do { fprintf(stderr, "%s\n", e); exit(EXIT_FAILURE); } while (0);

ssize_t push_through(const char *prefix,
        FILE *out, int in,
        char *buf, size_t bufsz)
{
    ssize_t bytes = read(in, buf, bufsz);

    if (bytes > 0)
        fprintf(out, "%s: (%.*s)\n", prefix, (int) bytes, buf);

    return bytes;
}

int main(void)
{
    pid_t pid;
    int stdout_pipe[2];
    int stderr_pipe[2];
    char buffer[4096];

    if (pipe(stdout_pipe) == -1 || pipe(stderr_pipe) == -1)
        die("pipe");

    if ((pid = fork()) == -1)
        die("fork");

    if (pid == 0) {
        close(stdout_pipe[0]);
        close(stderr_pipe[0]);

        dup2(stdout_pipe[1], STDOUT_FILENO);
        dup2(stderr_pipe[1], STDERR_FILENO);

        close(stdout_pipe[1]);
        close(stderr_pipe[1]);

        execl("/bin/lsx", "ls", "-1", (char *) NULL);
        perror("execl failed because");
        die("execl");
    } else {
        ssize_t outbytes, errbytes;

        close(stdout_pipe[1]);
        close(stderr_pipe[1]);

        do {
            outbytes = push_through("out", stdout, stdout_pipe[0],
                    buffer, sizeof buffer);
            errbytes = push_through("err", stderr, stderr_pipe[0],
                    buffer, sizeof buffer);
        } while (outbytes > 0 || errbytes > 0);

        wait(NULL);
    }
}

In the event that execl fails, it might be useful to call perror to print a more detailed error message.

Note that expecting the error

bash: /bin/lsx: no such file or directory

is possibly misguided, as exec* functions only defer to a shell under certain conditions.

The only mention of a shell in man 3 exec is with regards to execlp, execvp, and execvpe:

If the header of a file isn't recognized (the attempted execve(2) failed with the error ENOEXEC), these functions will execute the shell (/bin/sh) with the path of the file as its first argument. (If this attempt fails, no further searching is done.)