[1/7] qtest: don't use system command to avoid double fork

Submitted by Anthony Liguori on April 16, 2013, 2:45 p.m.

Details

Message ID 1366123521-4330-2-git-send-email-aliguori@us.ibm.com
State New
Headers show

Commit Message

Anthony Liguori April 16, 2013, 2:45 p.m.
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>
---
 tests/libqtest.c | 11 ++++-------
 1 file changed, 4 insertions(+), 7 deletions(-)

Patch hide | download patch | download mbox

diff --git a/tests/libqtest.c b/tests/libqtest.c
index 389596a..884f959 100644
--- 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);