Re: Help with break in printf function

[Andreas Schwab <schwab@linux-m68k.org>]

On Jan 26 2024, Carl Love wrote:

> I then single step to the print statement in main.
>
>
> gdb) s
> 23	  printf ("value is %d\n", value);
> (gdb) s
> Downloading source file /usr/src/debug/glibc-2.37-16.fc38.ppc64le/stdio-common/../sysdeps/ieee754/ldbl-128ibm-compat/ieee128-printf.c
> ___ieee128_printf (format=0x100008f0 "value is %d\n")                                                    
>     at ../sysdeps/ieee754/ldbl-128ibm-compat/ieee128-printf.c:29
> 29	  done = __vfprintf_internal (stdout, format, ap,
>  
> I can see that it stepped into an ieee128-printf.c file. But when I print out the location of the breakpoints
>
> (gdb) maint info breakpoints
> Num     Type           Disp Enb Address            What
> 1       breakpoint     keep y   0x00007ffff7a6f8b8 in __printf at printf.c:28 inf 1
> 1.1                         y   0x00007ffff7a6f8b8 in __printf at printf.c:28 inf 1
> 2       breakpoint     keep y   0x0000000010000788 in main 
>                                                    at /.../gdb/testsuite/gdb.base/annota1.c:15 inf 1
>
> I see the breakpoint is set in a different file printf.c.

That's because nothing in the program actually calls the printf
function, it calls ___ieee128_printf instead.  The function in printf.c
implements the pre-ieee128 printf function, and due to compiler magic
the printf call in the source is rewritten to __ieee128_printf.

> So, it would seem that GDB doesn't know that the test is compiled to use the print routine in ieee128-printf.c?

Probably GDB needs to implement some magic to know that printf and
__ieee128_printf are somewhat related.  Similar issues exist with
__isoc99_ and __isoc23_ prefixed symbols for all architectures.

-- 
Andreas Schwab, schwab@linux-m68k.org
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