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

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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
This commit is contained in:
Anthony Liguori 2013-04-16 09:45:15 -05:00
parent 4aead69241
commit 8a8fd63734
1 changed files with 4 additions and 7 deletions
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11
tests/libqtest.c
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@ -107,7 +107,7 @@ static pid_t qtest_qemu_pid(QTestState *s)
QTestState *qtest_init(const char *extra_args) QTestState *qtest_init(const char *extra_args)
{ {
QTestState *s; QTestState *s;
int sock, qmpsock, ret, i; int sock, qmpsock, i;
gchar *pid_file; gchar *pid_file;
gchar *command; gchar *command;
const char *qemu_binary; const char *qemu_binary;
@ -136,10 +136,8 @@ QTestState *qtest_init(const char *extra_args)
"%s", qemu_binary, s->socket_path, "%s", qemu_binary, s->socket_path,
s->qmp_socket_path, pid_file, s->qmp_socket_path, pid_file,
extra_args ?: ""); extra_args ?: "");
execlp("/bin/sh", "sh", "-c", command, NULL);
ret = system(command); exit(1);
exit(ret);
g_free(command);
} }
s->fd = socket_accept(sock); s->fd = socket_accept(sock);
@ -169,9 +167,8 @@ void qtest_quit(QTestState *s)
pid_t pid = qtest_qemu_pid(s); pid_t pid = qtest_qemu_pid(s);
if (pid != -1) { if (pid != -1) {
/* kill QEMU, but wait for the child created by us to run system() */
kill(pid, SIGTERM); kill(pid, SIGTERM);
waitpid(s->child_pid, &status, 0); waitpid(pid, &status, 0);
} }
unlink(s->pid_file); unlink(s->pid_file);