Re: Setting a floating point register to raw hex value

[Daniel Jacobowitz <drow@false.org>]

On Thu, Oct 18, 2007 at 06:16:00PM -0200, Thiago Jung Bauermann wrote:
> Hi,
> 
> Yes, that's indeed crazy. :-)
> 
> Just found out that this works:
> 
> (gdb) set (void *) $f1 = 0x1
> (gdb) info reg f1
> f1             4.9406564584124654e-324  (raw 0x0000000000000001)

I think this only works by an accident involving
gdbarch_convert_register_p on PowerPC.  We should find some proper way
to do it, document that, and then make this not work :-)

> On Thu, 2007-10-18 at 15:38 -0400, Grzegorz Cieslewski wrote:
> > I have figured out a workaround but it is really ugly.
> > 
> > Step 1:  Create shared object containing a union as follows
> > union longlong2double
> > {
> > long long int hexNum;
> > double floatNum;
> > } temp;
> > 
> > Step 2: Force load the library at runtime into your program using
> > LD_PRELOAD environmental variable
> > 
> > Step 3: When debugging the program store 0x1 into temp.hexNum;
> > (one would think that I could use the following now 'set $f0 =
> > temp.floatNum; but that yields incorrect result)
> > 
> > Step4:  Overwrite the current instruction with 'load float'
> > 
> > Step5: Using stepping command 'stepi 1' to write the value to the $f0
> > 
> > Step6: Undo all the changes required by steps 4 and 5
> > 
> > Does any one know of a simpler way because this workaround is crazy?

Sure, here's a much easier way that ought to work:

(gdb) set {long long} ($sp - 8) = 1
(gdb) set $f0 = {double} ($sp - 8)

Doesn't have to be $sp - 8, any memory address will do.

Oddly, this does not work on PowerPC.  I can't figure out why
not.  It certainly ought to work; perhaps it requires a current
version of GDB, since 0x1 is a denormal.  It does work for normals,
though, so I know the approach is sound.

-- 
Daniel Jacobowitz
CodeSourcery