$pid = fork;
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Congratulations, now you have two processes. And they both will execute the next block:

      eval {
        alarm $timeout;
        $result = `$solver $file`;
        waitpid($pid, 0);
      };
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So now you have two solver programs.

The standard idiom for forking goes something like this:

my $pid = fork();
die "failed to fork" unless defined $pid;
if ($pid) {
  # we are parent. do things
} else {
  # we are child. do other things
}
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You probably want to put your eval block into the child portion. However -- this still won't work, as the child cannot return the $result variable to the parent -- they don't share memory. Returning a value will require some ipc trickery.

But the trickery is not worth it. Just listen to the other poster and don't fork. It'll work fine.