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 2bbc5b43b28eca1ff941052f8ddc6602e2b7b2aa..da0d8d3f4214fb298e1451eb35fb387c083ef6a0 100644 (file)
--- a/src/bin/lttng-sessiond/ust-consumer.c
+++ b/src/bin/lttng-sessiond/ust-consumer.c
@@ -380,8 +380,8 @@ 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.");
                }