Re: Simple x86 assembler question

2001-10-29 Thread John Baldwin
On 29-Oct-01 David O'Brien wrote: > On Sun, Oct 28, 2001 at 09:21:33AM -0700, John Baldwin wrote: >> Almost. The '2' there is a multiplier on (I think) %eax, so it uses >> 'ebx + 2 * eax + 0xe90' for the memory address. Either that or 'eax + >> 2 * ebx + 0xe90'. Check the gas info page for the

Re: Simple x86 assembler question

2001-10-28 Thread David O'Brien
On Sun, Oct 28, 2001 at 09:21:33AM -0700, John Baldwin wrote: > Almost. The '2' there is a multiplier on (I think) %eax, so it uses > 'ebx + 2 * eax + 0xe90' for the memory address. Either that or 'eax + > 2 * ebx + 0xe90'. Check the gas info page for the AT&T syntax to > figure out exactly whi

RE: Simple x86 assembler question

2001-10-28 Thread John Baldwin
On 27-Oct-01 Matthew Emmerton wrote: > Hi all, > > This weekend I decided to do some assembly hacking on some object-only code > that I've lost the C source for. Since I haven't coded assembler for at > least 8 years, and I threw my x86 assembly manuals out when I moved 6 months > ago, there ar

Re: Simple x86 assembler question

2001-10-28 Thread Seth Kingsley
On Sat, Oct 27, 2001 at 07:08:58PM -0400, Matthew Emmerton wrote: > In particular, am I interpreting these instructions correctly? > > 0x80839fb : movzbl (%edx,%eax,1),%eax > > > Takes %eax + %edx, obtains the byte value in memory at that address, > zero-extends and places into %eax > >

Re: Simple x86 assembler question

2001-10-28 Thread Sergey Babkin
Matthew Emmerton wrote: > > Hi all, > > This weekend I decided to do some assembly hacking on some object-only code > that I've lost the C source for. Since I haven't coded assembler for at > least 8 years, and I threw my x86 assembly manuals out when I moved 6 months > ago, there are a few thi