Old school ascii shellcode -


this part of ascii shellcode set eax @ 0 , instruction (%...). in debugger , in pratice pic code work time why? , instruction algorithme is:

operand target ← operand target ∩ operande source flag cf ← 0 flag of ← 0

it's possible eax register not set 0 if previous value of eax not good?

#include <stdlib.h>  int main() {     asm("\         , eax, 0x454e4f4a;\         , eax, 0x3a313035;\         ");      return 0; }

compilation line:

gcc -m32  -w -wall -std=gnu99 -masm=intel -g eaxzero.c -o eaxzero

gdb instruction:

    (gdb)   b 5 ..     (gdb)   r ..     (gdb)   x/i $eip ..     (gdb)   x/x $eax ..     (gdb)   nexti ..     (gdb)   x/x $eax ..     (gdb)   nexti ..     (gdb)   x/x $eax ..


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