Re: Shouldn't gdb print the components of the function descriptor when you print a pointer to a function?

[Andrew Cagney]

> So my problem was caused by the function descriptor not being visible
> when I printed the pointer to the function with gdb. Why isn't gdb being
> fixed so that the function descriptor is visible?


While I agree that being able to display the descriptor would be a 
useful feature, implementing it isn't that straight forward.  It leads 
to all sort of edge conditions, I'll note three of them below.

Can I suggest sending further comments to address@hidden or 
recording the feature request with a bug report in 
http://sources.redhat.com/gdb/bugs:

enjoy,
Andrew

--

The descriptor isn't visible from C so there is no C construct available 
 that lets you access it.  vis:

(top-gdb) print main
$1 = {int (int, char **)} 0x1803458 <main>
(top-gdb) print &main
$2 = (int (*)(int, char **)) 0x1803458 <main>
(top-gdb) print *(&main)
$3 = {int (int, char **)} 0x1803458 <main>

A mechanism outside of the normal C expression / type system would be 
needed.

--

Adding to the fun, consider the code:

        extern void foo (void);
        long bar = &foo;

and the GDB sequence:

        (gdb) print bar
        (gdb) print &foo

For GDB to print identical values for ``bar'' and ``&foo'' it needs to 
find, in target memory, the descriptor for ``&foo'' that GCC created. 
GDB does try to do this.

--

Finally, GCC has this wonderful feature (not) where the user can specify:

        foo + 10

GCC converts it to something like:

        &foo + 10

For a descriptor based ABI, is turns out to be pretty meaningless.  It 
is a random pointer into target data space.  GDB doesn't emulate this 
since it leads to commands such as:

        (gdb) disassemble foo foo + 10

being meaningless.

.