NASM彙編數轉換爲

Question

我對我來說似乎是一個相當直接的項目。詢問用戶0-255之間的數字和2-9之間的數字。並在該基地輸出。我打算簡單地執行通常的除法算法並獲取剩餘部分，推入堆棧，然後退出以獲取相反的順序以輸出回用戶。但是將幾個打印調試語句後，我得到很奇怪的輸出剩餘

; Run using nasm -f elf -g -F stabs proj4.asm ; gcc -o proj4 proj4.o -m32 ; To execute type proj4 %macro SAVE_REGS 0 push eax push ecx push edx %endmacro %macro RESTORE_REGS 0 pop edx pop ecx pop eax %endmacro %macro CALL_PUTS 1 push %1 call puts add esp, 4 %endmacro %macro CALL_SCANF 2 push %1 push %2 call scanf add esp, 8 %endmacro %macro CALL_PRINTF1 1 push %1 ;The address of the string to print call printf add esp, 4 %endmacro %macro CALL_PRINTF2 2 push %1 ;The formatted string with a %char place holder push %2 ;The item to place into the place holder call printf add esp, 8 %endmacro SECTION .data prmptNumMsg: db "Enter a number between 0 and 255: ", 0 prmptBaseMsg: db "Enter a base between 2 and 9: ", 0 remShow: db 'The the remainder is: %d', 10, 0 numShow: db 'The number is %d', 10, 0 baseShow: db 'The base is %d', 10, 0 printed: db 'Looped in division method', 10, 0 poped: db 'Looped in pop method',10, 0 ansShow: db '8d', 10, 0
numFormat db '%d', 0 stringFormat db '%s', 0 SECTION .bss numVal resd 1 baseVal resd 1 ans resd 9 i resd 1 n resd 1 j resd 1

SECTION .text global main extern puts extern scanf extern printf main: push ebp ; Set up stack frame for debugger mov ebp, esp push ebx push esi push edi ;Everything before this is boilerplate
getInt: CALL_PRINTF1 prmptNumMsg ;push numVal ;push numFormat ;call scanf ;add esp, 8 CALL_SCANF numVal, numFormat
mov eax, dword[numVal] mov ebx, dword 0 ;check below 0 cmp eax, ebx jb getInt mov eax, dword[numVal] mov ebx, dword 255 ;check above 255 cmp eax , ebx ja getInt ;VALID INTEGER PAST THIS POINT
getBase: CALL_PRINTF1 prmptBaseMsg
CALL_SCANF baseVal, numFormat mov eax, dword[baseVal] mov ebx, dword 0 cmp eax, ebx jb getBase mov eax, dword[baseVal] mov ebx, dword 9 cmp eax, ebx ja getBase ;END GETBASE ;VALID BASE NUMBER PAST THIS POINT mov eax, dword[numVal] mov [n], eax ;set n to the current number value CALL_PRINTF2 eax, numShow mov eax, dword 0 mov [i], eax mov eax, dword[baseVal] CALL_PRINTF2 eax, baseShow doDivision: ;CALL_PRINTF1 printed xor edx, edx mov eax, dword[n] mov ebx, dword[baseVal] div ebx ;edx = remainder eax = quotient mov [n], eax ;save quotient
CALL_PRINTF2 eax, numShow CALL_PRINTF2 edx, remShow
;push edx ;save remainder on stack to pop in reverse order later ;mov [n], eax ;move quotient to eax mov ebx, dword[i] inc ebx mov [i], ebx ;i++ ;mov eax, [i] mov ecx, dword 8 cmp ebx, ecx jb doDivision ;END DO DIVISION

end: ;Everything after this is boilerplate pop edi pop esi pop ebx mov esp, ebp pop ebp ret

當程序運行我的輸出如下 Enter a number between 0 and 255: 105 Enter a base between 2 and 9: 4 The number is 105 The base is 4 The number is 26 The the remainder is: 13144896 The number is 6 The the remainder is: 13144896 The number is 1 The the remainder is: 13144896 The number is 0 The the remainder is: 13144896 The number is 0 The the remainder is: 13144896 The number is 0 The the remainder is: 13144896 The number is 0 The the remainder is: 13144896 The number is 0 The the remainder is: 13144896 我希望它環路8倍這是正確的，每一個DIV操作的商是正確的，但我得到瘋狂的數字爲剩下這將在edx舉行。我不明白是什麼造成這種情況。

Answer 1

當你調用一個例程（例如：printf）時，有一些規則管理這個工作原理。例如，你把參數放在棧上嗎？或者將它們傳入寄存器？你是否從左向右推參數？或者從左到右？調用者是否將參數從堆棧中彈出？還是被調用者？返回值的位置在哪裏？

最重要的是對於你的問題，被調用者是否需要確保所有寄存器在返回時都具有相同的值？或者它可以覆蓋其中的一些？

這些問題的答案被稱爲「調用約定」（或有時ABI）。並不只有一個答案。例如，cdecl，pascal和fastcall都是x86上的常用調用約定，它們都以稍微不同的方式回答這些問題。

我相信你會發現printf是cdecl。考慮到這一點，你可以檢查出http://en.wikipedia.org/wiki/X86_calling_conventions#cdecl。這應該有助於你理解在這個例子中edx發生了什麼。