NASM部件号到基本的转换

Question

我觉得这是一个相当简单的项目。 要求用户输入一个介于0-255和2-9之间的数字。 并在该基础上输出。 我打算简单地执行通常的除法算法，并取余数，推入堆栈，然后弹出以得到相反的顺序以输出给用户。 但是，在插入一些打印调试语句后，剩下的输出非常奇怪

; Run using nasm -f elf -g -F stabs proj4.asm ; gcc -o proj4 proj4.o -m32 ; To execute type proj4 ; 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 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 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 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 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 ;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

运行程序时，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我希望它循环8次是正确的，每个div操作的商是正确的，但是我为要保留的余数得到了疯狂的数字在edx中。 我不明白是什么原因造成的。

Answer 1

当您调用例程（例如：printf）时，有一些规则来控制其工作方式。 例如，您是否将参数放在堆栈上？ 还是将它们传递给寄存器？ 您是否从左向右推送参数？ 还是从右到左？ 调用方是否将参数弹出堆栈？ 还是被呼叫者？ 返回值将位于何处？

对于您的问题最重要的是，被调用方是否需要确保所有寄存器在返回时都具有相同的值？ 还是可以覆盖其中的一些？

这些问题的答案称为“呼叫约定”（有时称为ABI）。 而且，答案不只一个。 例如，cdecl，pascal和fastcall都是x86上的常见调用约定，它们都以略有不同的方式回答这些问题。

我相信您会发现printf是cdecl。 考虑到这一点，您可以查看http://en.wikipedia.org/wiki/X86_calling_conventions#cdecl 。 这应该可以帮助您了解本示例中edx发生了什么。