reversal of fsubp and fsubrp

[dancie]

I have noticed when debugging with gdb that the opcode "de e9" that (in intel
syntax) gdb gives out fsubrp st1,st0.
But according to the intel docs this fsubp st1,st0. I have also noticed that
the opcode for fsubrp st1,st0 gives out
fsubp st1,st0.  I have not yet tried the other reverse pop instructions but
these two are switched. Here is a sample from
objdump output which gives an example:
000004b0 <etyBit>:
 4b0:   ba 01 00 00 00          mov    $0x1,%edx
 4b5:   dd d8                   fstp   %st(0)
 4b7:   dd d8                   fstp   %st(0)
 4b9:   d9 ee                   fldz   
 4bb:   dd dc                   fstp   %st(4)
 4bd:   db 2d 00 00 00 00       fldt   0x0
 4c3:   db 3d 00 00 00 00       fstpt  0x0
 4c9:   db 2d 00 00 00 00       fldt   0x0
 4cf:   db 3d 00 00 00 00       fstpt  0x0
 4d5:   db 2d 00 00 00 00       fldt   0x0
 4db:   db 2d 00 00 00 00       fldt   0x0
 4e1:   de e9                   fsubrp %st,%st(1)   <---- Here is the error
 4e3:   da 35 00 00 00 00       fidivl 0x0
 
I first thought it was an error in gdb but a received an email from Daniel
Jacobowitz, from gdb. He said that is was
possibly an error in binutils.

Dancie Reeves

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