Re: Getting pissed off by gdb. Please help with stepping in.

[Doug Evans <dje@google.com>]

To me it's the difference between stepping during

  foo(1);

and

  foo (bar());

In the former, the last "step" that returns from foo returns to the
caller - it's intuitive and what I expect.
In the latter, the last "step" in bar doesn't return to the caller,
instead it proceeds until we get into foo.  I just stepped out of a
function - how come I've also stepped into another function?

Also, when typing "s" and then constantly hitting <return> to move
things along, it's useful to see that extra step in the flow of the
program: the point between the return from bar and the call to foo.

As a data point, there is a time where stepping does stop in the
middle of a source line: next over longjmp (he says after having just
revisited why longjmp.exp has failures on amd64 (pointer mangling) and
tried it on i386).  According to the thread from bug 9270, the
intended fix IIUC is to keep stepping through the longjmp until it
returns - I wonder if there's an overlap here.

On Thu, Mar 18, 2010 at 11:37 AM, Paul Koning <Paul_Koning@dell.com> wrote:
> I'd say it is useful because it matches the documentation.  "s" is
> documented as "run until you're at a different source line".  In
> foo(bar()), the source line you come to after exit from bar() is the
> call to foo(), not the first line of foo -- that would take "s 2".
>
>        paul
>
>> -----Original Message-----
>> From: gdb-owner@sourceware.org [mailto:gdb-owner@sourceware.org] On
>> Behalf Of Eli Zaretskii
>> Sent: Thursday, March 18, 2010 2:31 PM
>> To: Doug Evans
>> Cc: temp@sourceboost.com; gdb@sourceware.org
>> Subject: Re: Getting pissed off by gdb. Please help with stepping in.
>>
>> > Date: Thu, 18 Mar 2010 08:10:41 -0700
>> > From: Doug Evans <dje@google.com>
>> > Cc: temp@sourceboost.com, gdb@sourceware.org
>> >
>> > Here's what gdb 7.1 does:
>> >
>> > (gdb) f
>> > #0  bar () at stepout.c:4
>> > 4       int bar () { return 1; }
>> > (gdb) s
>> > foo (x=1) at stepout.c:6
>> > 6       void foo (int x) { g = x; }
>> > (gdb)
>> >
>> > Note that we've stepped out of bar and into foo.
>> >
>> > Here is what Pavel is expecting instead:
>> >
>> > (gdb) f
>> > #0  bar () at stepout.c:4
>> > 4       int bar () { return 1; }
>> > (gdb) s
>> > 0x00000000004003b7 in main () at stepout.c:11
>> > 11        foo (bar ());
>> > (gdb)
>>
>> Thanks.  But why is ``what Pavel is expecting'' useful?  What use-case
>> does it handle that the v7.1 behavior does not?
>