Hi,
If you see a XXX'N in callgrind's output, this means that this is the
profile for all invocations of function XXX at recursion level N. For
level 1, suffix '1 is omitted. And there is one caveat: if you ask
callgrind to only distinguish n levels, XXX'n will include all
XXX'm with m > n.
If you run "valgrind --tool=callgrind -h", you see in the help output:
...
--separate-callers=<n> Separate functions by call chain length [0]
--separate-callers<n>=<f> Separate <n> callers for function <f>
--separate-recs=<n> Separate function recursions up to level [2]
--separate-recs<n>=<f> Separate <n> recursions for function <f>
...
These are callgrind's options which control what to do about recursive
invocations. The default is that 2 levels are distinguished. The means
for a function XXX, the profile may have info about XXX and XXX'2, with
XXX'2 including all recursion levels above 1.
Another way to distinguish invocations of the same function is to
add the call chain as context ("--separate-callers" above). E.g. setting
this to 2 adds a call chain of up to length 2 as context to a function
symbol: XXX'Y'Z would mean all invocations of XXX only when directly
called from Y which is called from Z.
Distinguishing function invocations is useful for the visualization
(k/qcachegrind) not to have to group all functions in recursive cycles
together. Such cycle detection is done offline in the visualization.
You do not really want cycles as they obfuscates profile results:
inclusive cost is meaningless among functions in a cycle, so the
visualization has to suppress it.
Hope this helps,
Josef
Am 09.06.2016 um 17:13 schrieb Yang Liu:
> Hi, guys!
>
> I'm profiling postgreSQL with callgrind. Basically, there's a call chain
> in the following order.
>
> ExecProcNode->ExecAgg->ExecProcNode->ExecSeqScan
>
> ExecProcNode is called recursively through intermediate functions. And
> the output file marked the second call to ExecProcNode as a call to
> ExecProcNode'2.
>
> I read the manual, mailing list archive and callgrind's output format.
> But that doesn't help me.
> Cheers,
> Yang Liu
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