Hello, for learning about gdb macros I looked through gdbinit-1.i386. ( or is there some secret gdb manual that *explains* macros? :-)
The macros xp and xxp calculate the number of possible arguments as: ((*(int*)$ebp)-(int)$ebp)/4-4 Let's see (assuming "char* ebp"): *ebp : saved ebp from previous frame *(ebp+4) : saved eip *(ebp+8): first parameter (if any) *((*ebp)-4): saved register or local variable from previous frame Right? What I do not understand is the subtraction of 4 (I expected 2). When disassembling through different portions of the kernel code and userland programs, I found pushing of none up to three extra register (ebx, esi, ebp). But that depends heavyly on compiler options, doesn't it? So, is it just a guess that there are two register variables or am I missing something? Any comments? Norbert _______________________________________________ freebsd-hackers@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-hackers To unsubscribe, send any mail to "[EMAIL PROTECTED]"
