[英]How can I call this x86 ASM CALL in C++ with typedef or inline
[英]Call not returning properly [X86_ASM]
這是使用x86內聯匯編的C ++ [Intel語法]
功能:
DWORD *Call ( size_t lArgs, ... ){
DWORD *_ret = new DWORD[lArgs];
__asm {
xor edx, edx
xor esi, esi
xor edi, edi
inc edx
start:
cmp edx, lArgs
je end
push eax
push edx
push esi
mov esi, 0x04
imul esi, edx
mov ecx, esi
add ecx, _ret
push ecx
call dword ptr[ebp+esi] //Doesn't return to the next instruction, returns to the caller of the parent function.
pop ecx
mov [ecx], eax
pop eax
pop edx
pop esi
inc edx
jmp start
end:
mov eax, _ret
ret
}
}
此功能的目的是調用多個功能/地址,而無需單獨調用它們。
為什么我要調試它? 我白天必須上學,晚上需要上學。
非常感謝,iDomo
感謝您提供一個完整的可編譯示例,它使解決問題變得更加容易。
根據您的Call
函數簽名,當設置了堆棧幀時, lArgs
位於ebp+8
,並且指針開始於ebp+C
還有其他一些問題。 這是經過修正的版本,具有一些推送/彈出優化和清理功能,已在MSVC 2010(16.00.40219.01)上進行了測試:
DWORD *Call ( size_t lArgs, ... ) {
DWORD *_ret = new DWORD[lArgs];
__asm {
xor edx, edx
xor esi, esi
xor edi, edi
inc edx
push esi
start:
cmp edx, lArgs
; since you started counting at 1 instead of 0
; you need to stop *after* reaching lArgs
ja end
push edx
; you're trying to call [ebp+0xC+edx*4-4]
; a simpler way of expressing that - 4*edx + 8
; (4*edx is the same as edx << 2)
mov esi, edx
shl esi, 2
add esi, 0x8
call dword ptr[ebp+esi]
; and here you want to write the return value
; (which, btw, your printfs don't produce, so you'll get garbage)
; into _ret[edx*4-4] , which equals ret[esi - 0xC]
add esi, _ret
sub esi, 0xC
mov [esi], eax
pop edx
inc edx
jmp start
end:
pop esi
mov eax, _ret
; ret ; let the compiler clean up, because it created a stack frame and allocated space for the _ret pointer
}
}
完成后,別忘了delete[]
從此函數返回的內存。
我注意到,在調用之前,您按順序按了EAX,EDX,ESI,ECX,但返回后不要以相反的順序彈出。 如果第一個CALL正確返回,但隨后的CALL沒有返回,則可能是問題所在。
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