On Thu, 25 Jan 2024 11:28:34 -0500 Steven Rostedt <rostedt@xxxxxxxxxxx> wrote: > On Thu, 25 Jan 2024 16:16:42 +0100 > Pierre Gondois <pierre.gondois@xxxxxxx> wrote: > > > >> For example, with a trace recorded with: > > >> $ trace-cmd record -e sched_wakeup -B switch_instance \ > > >> -e sched_switch -B timer_instance -e timer > > > > > > I tried this with a trace.dat file that has a instance called "tast" and it > > > didn't work. Note, the top instance had no data. > > > > > > $ trace-cmd split -i /work/traces/trace-tast.dat -o /tmp/trace-tast2.dat -B tast > > > > With this command, the expected behaviour should be: > > trace-tast2.dat: > > - Contains all the instances of the input trace-tast.dat file > > Wait what? -i should *not* be used for output if -o is specified! OK, I see what you did. As if I wrote: trace-cmd split -i /work/traces/trace-tast.dat -B tast -o /tmp/trace-tast2.dat I got what I want. But once -o is used, it should repeat for the -B. the -B should *not* use the input file unless -o is not specified. trace-cmd split -i /work/traces/trace-tast.dat -o /tmp/trace-tast2.dat -B tast What the above should have done the same. But if we added: trace-cmd split -i /work/traces/trace-tast.dat --top -o /tmp/trace-tast2.dat -B tast Then /tmp/trace-tast2.dat would just have the top instance, and it should create: /tmp/trace-tast2.dat.tast to hold the tast image. Now if I had had: trace-cmd split -i /work/traces/trace-tast.dat --top -B tast -o /tmp/trace-tast2.dat Then because there was a command "--top" before the -B but it had no -o assigned for it, then the input_file would be considered the output file and that should act like you described. -- Steve
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