qtest: don't use system command to avoid double fork

Currently we waitpid on the child process we spawn off that does
nothing more than system() another process.  While this does not
appear to be incorrect, it's wasteful and confusing so get rid of
it.

Signed-off-by: Anthony Liguori <aliguori@us.ibm.com>
Message-id: 1366123521-4330-2-git-send-email-aliguori@us.ibm.com

diff --git a/tests/libqtest.c b/tests/libqtest.c
index 389596a..884f959 100644 (file)
--- a/tests/libqtest.c
+++ b/tests/libqtest.c
@@ -107,7 +107,7 @@ static pid_t qtest_qemu_pid(QTestState *s)
 QTestState *qtest_init(const char *extra_args)
 {
     QTestState *s;
-    int sock, qmpsock, ret, i;
+    int sock, qmpsock, i;
     gchar *pid_file;
     gchar *command;
     const char *qemu_binary;
@@ -136,10 +136,8 @@ QTestState *qtest_init(const char *extra_args)
                                   "%s", qemu_binary, s->socket_path,
                                   s->qmp_socket_path, pid_file,
                                   extra_args ?: "");
-
-        ret = system(command);
-        exit(ret);
-        g_free(command);
+        execlp("/bin/sh", "sh", "-c", command, NULL);
+        exit(1);
     }
 
     s->fd = socket_accept(sock);
@@ -169,9 +167,8 @@ void qtest_quit(QTestState *s)
 
     pid_t pid = qtest_qemu_pid(s);
     if (pid != -1) {
-        /* kill QEMU, but wait for the child created by us to run system() */
         kill(pid, SIGTERM);
-        waitpid(s->child_pid, &status, 0);
+        waitpid(pid, &status, 0);
     }
 
     unlink(s->pid_file);