NASM汇编数转换为

Question

我对我来说似乎是一个相当直接的项目。询问用户0-255之间的数字和2-9之间的数字。并在该基地输出。我打算简单地执行通常的除法算法并获取剩余部分，推入堆栈，然后退出以获取相反的顺序以输出回用户。但是将几个打印调试语句后，我得到很奇怪的输出剩余

; 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

当程序运行我的输出如下 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 我希望它环路8倍这是正确的，每一个DIV操作的商是正确的，但我得到疯狂的数字为剩下这将在edx举行。我不明白是什么造成这种情况。

Answer 1

当你调用一个例程（例如：printf）时，有一些规则管理这个工作原理。例如，你把参数放在栈上吗？或者将它们传入寄存器？你是否从左向右推参数？或者从左到右？调用者是否将参数从堆栈中弹出？还是被调用者？返回值的位置在哪里？

最重要的是对于你的问题，被调用者是否需要确保所有寄存器在返回时都具有相同的值？或者它可以覆盖其中的一些？

这些问题的答案被称为“调用约定”（或有时ABI）。并不只有一个答案。例如，cdecl，pascal和fastcall都是x86上的常用调用约定，它们都以稍微不同的方式回答这些问题。

我相信你会发现printf是cdecl。考虑到这一点，你可以检查出http://en.wikipedia.org/wiki/X86_calling_conventions#cdecl。这应该有助于你理解在这个例子中edx发生了什么。