RE: [Mingw-users] How to read a map file?

[Dave H. <dhy...@br...>]

Hi Greg, Semmler,

...snip...
> Here's a piece of a map file I have:
>=20
>   .data          0x004c9000       0x10=20
> C:/MinGW/bin/../lib/gcc-lib/mingw32/3.2.3/../../../crt2.o
>                  0x004c9000                _argc
>                  0x004c9004                _argv
>   .data          0x004c9010       0x10=20
> C:/local/lib/libmpatrol.a(malloc.o)

The 0x10 over on the right tells you how big the portion of .data is
from crt2.o. Note that trying to calculate symbol size is kind of dodgy.
This is because you're only seeing the public symbols. Variables which
are declared using static take up space but don't show in the map file.

> The .data section seems to be sorted by address. I think we can
> conclude from this map that main()'s arguments start at the
> address given, and that _argc probably takes four bytes, and
> that _argv takes no more than twelve bytes, though it probably
> takes less. I'm guessing that the difference between successive
> addresses reflects not just object size but also alignment.

Yes, alignment will be absorbed into your variables. If you produce the
assembly (using mingw32-gcc -S), you'll see something like this:

.globl _z
   .data
   .align 4
_z:
   .long 9

(The C code was "int z =3D 9;"), so 16 byte alignment is being used. For
i386, the ".align n" means 2 to the power n. Some processors interpret
".align n" as align to an n byte boundary.

> I'm casually assuming that the authors consider the map layout
> self-explanatory, but if you find a good document that explains
> this in detail, I'd like to hear about it.
>=20
> Maybe something like 'objdump' or 'size' would be more useful
> to you. Last time I checked (years ago) they were documented in
> the 'binutils' manual.

Yep - still are.

--
Dave Hylands
Vancouver, BC, Canada
http://www.DaveHylands.com/=20