execvp（）之后cout / stdout不起作用

Question

So I am attempting to use execvp() from unistd.h in a short program I'm writing. However for some strange reason I seem to be losing the ability to use cout or even printf after calling the execvp function.

For example this works:

pid_t pid;
int status;
if ((pid = fork()) > 0) {
  waitpid(pid, &status, 0);
}
else {
  execvp(argv[1], &argv[1]);
}
cout << "DONE!" << endl;

But this does not work:

execvp(argv[1], &argv[1]);
cout << "DONE!" << endl;

And, while it is not a huge problem, I would like to understand why it is happening. I have not been able to find anything relevant when I searched here and with Google.

Not sure if this has anything at all to do with the issue but I am using the -std=c++11 flag with g++ .

Answer 1

This will not work

execvp(argv[1], &argv[1]);
cout << "DONE!" << endl;

Because the function execvp() never returns (if it succeeded).

It replaces the current processes with a new image and executes that.

On the other this works:

if ((pid = fork()) > 0) {
  waitpid(pid, &status, 0);
}
else {
  execvp(argv[1], &argv[1]);
}
cout << "DONE!" << endl;

What happens here is that fork() creates a new processes. So you now have two processes at exactly the same place.

• One processes has the variable pid set to zero (this is the parent). It goes into the first branch of the if statement and there waits for the child to finish.
• One processes has the variable pid set to none zero (this is the child). It goes into the second branch. It executes execvp() which never returns. It does not return because the processes image is replaced by another executable.

Note what you should really do is:

if ((pid = fork()) > 0) {
  waitpid(pid, &status, 0);
  cout << "Child Finished!" << endl;
}
else {
  execvp(argv[1], &argv[1]);
  // This code should never execute if everything is OK.
  cout << "Child failed to start" << endl;
  exit(1);
}