Fix: don't print stream name in error message

The stream received, in per UID, is actually a temporary stream object
that only contains the UST object data which is the relevant part for
UST to use.

Thus on error the name was random data thus print the valid handle
descriptor instead of invalid data.

Signed-off-by: David Goulet <dgoulet@efficios.com>

diff --git a/src/bin/lttng-sessiond/ust-consumer.c b/src/bin/lttng-sessiond/ust-consumer.c
index 9715c426f468f5b24cf76bf64726b0cf218d5af3..a8a11dcee41729f56da30f1f34e62b0a7fa3b4b2 100644 (file)
--- a/src/bin/lttng-sessiond/ust-consumer.c
+++ b/src/bin/lttng-sessiond/ust-consumer.c
@@ -383,8 +383,8 @@ int ust_consumer_send_stream_to_ust(struct ust_app *app,
        ret = ustctl_send_stream_to_ust(app->sock, channel->obj, stream->obj);
        if (ret < 0) {
                if (ret != -EPIPE && ret != -LTTNG_UST_ERR_EXITING) {
-                       ERR("Error ustctl send stream %s to app pid: %d with ret %d",
-                                       stream->name, app->pid, ret);
+                       ERR("ustctl send stream handle %d to app pid: %d with ret %d",
+                                       stream->obj->handle, app->pid, ret);
                } else {
                        DBG3("UST app send stream to ust failed. Application is dead.");
                }