sched/core: Convert get_task_struct() to return the task

Returning the pointer that was passed in allows us to write
slightly more idiomatic code.  Convert a few users.

Signed-off-by: Matthew Wilcox (Oracle) <willy@infradead.org>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Link: https://lkml.kernel.org/r/20190704221323.24290-1-willy@infradead.org
Signed-off-by: Ingo Molnar <mingo@kernel.org>

1 files changed, 3 insertions, 6 deletions

diff --git a/kernel/events/core.c b/kernel/events/core.c
index 026a14541a38..ea5e8139fe62 100644
--- a/kernel/events/core.c
+++ b/kernel/events/core.c
@@ -4089,10 +4089,8 @@ alloc_perf_context(struct pmu *pmu, struct task_struct *task)
                 return NULL;
 
         __perf_event_init_context(ctx);
-        if (task) {
-                ctx->task = task;
-                get_task_struct(task);
-        }
+        if (task)
+                ctx->task = get_task_struct(task);
         ctx->pmu = pmu;
 
         return ctx;
@@ -10355,8 +10353,7 @@ perf_event_alloc(struct perf_event_attr *attr, int cpu,
                  * and we cannot use the ctx information because we need the
                  * pmu before we get a ctx.
                  */
-                get_task_struct(task);
-                event->hw.target = task;
+                event->hw.target = get_task_struct(task);
         }
 
         event->clock = &local_clock;