Assembly io.h出错有什么问题？

这是我课本上直接写的，不知道是什么原因造成的。它以前工作过

我正在使用的图书代码：

; Input x and y, call procedure to evaluate 3*x+7*y, display result
; Author:  R. Detmer
; Date:    6/2013
.586
.MODEL FLAT
INCLUDE io.h
.STACK 4096

.DATA
number1 DWORD   ?
number2 DWORD   ?
prompt1 BYTE    "Enter first number x", 0
prompt2 BYTE    "Enter second number y", 0
string  BYTE    20 DUP (?)
resultLbl BYTE  "3*x+7*y", 0
result  BYTE    11 DUP (?), 0

.CODE
_MainProc PROC
        input   prompt1, string, 20      ; read ASCII characters
        atod    string          ; convert to integer
        mov     number1, eax    ; store in memory

        input   prompt2, string, 20      ; repeat for second number
        atod    string
        mov     number2, eax

        push    number2         ; 2nd parameter
        push    number1         ; 1st parameter
        call    fctn1           ; fctn1(number1, number2)
        add     esp, 8          ; remove parameters from stack

        dtoa    result, eax     ; convert to ASCII characters
        output  resultLbl, result  ; output label and result

        mov     eax, 0  ; exit with return code 0
        ret
_MainProc ENDP

; int fctn1(int x, int y)
; returns 3*x+7*y
fctn1   PROC
        push    ebp             ; save base pointer
        mov     ebp, esp        ; establish stack frame
        push    ebx             ; save EBX

        mov     eax, [ebp+8]    ; x
        imul    eax, 3          ; 3*x
        mov     ebx, [ebp+12]   ; y
        imul    ebx, 7          ; 7*y
        add     eax, ebx        ; 3*x + 7*y

        pop     ebx             ; restore EBX
        pop     ebp             ; restore EBP
        ret                     ; return      
fctn1   ENDP

END

错误：

Error   101 error A1012: error count exceeds 100; stopping assembly C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\io.h 102 1   console32
Error   3   error A2008: syntax error : *   C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\io.h 3   1   console32
Error   13  error A2044: invalid character in file  C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\include\io.h 14  1   console32
Error   102 error MSB3721: The command "ml.exe /c /nologo /Zi /Fo"Debug\Source.obj" /W3 /errorReport:prompt /Fl /TaSource.asm" exited with code 1.  C:\Program Files (x86)\MSBuild\Microsoft.Cpp\v4.0\V110\BuildCustomizations\masm.targets 49  5   console32

我去掉了重复的错误，使它看起来更干净/整洁。 我试图编辑头文件以删除注释，但我没有写入头文件的权限。我也不认为有必要编辑这个文件，因为它曾经工作过。我正在使用书籍、文件和代码

我在我的笔记本电脑上也试过了，它也给了我同样的错误。我在两台机器上都使用visual studio 2012

如果你需要更多信息，请告诉我

在windows32汇编程序中使用io.h。 打开VS-->新项目-->c/c++-->Windows32（非控制台32）-->空白项目。 在“解决方案资源管理器”中，右击“源文件夹”> >添加新项-> C++文件（.CPP）并用扩展名.ASM而不是.CPP来指定您的名称。 代码：

在windows32汇编程序中使用io.h。 打开VS-->新项目-->c/c++-->Windows32（非控制台32）-->空白项目。 在“解决方案资源管理器”中，右击“源文件夹”> >添加新项-> C++文件（.CPP）并用扩展名.ASM而不是.CPP来指定您的名称。 代码：

---

我道歉！我刚刚意识到我不能使用io.h，因为我正在编写一个控制台应用程序，这段代码是针对windows应用程序的

---

对于其他有此问题的人，如果您正在编写控制台程序，您不能使用io.h，控制台程序中没有输入/输出，如果您需要输入/输出，您必须使用windows应用程序。

@RossRidge对此我不确定，我知道它是用C编写的。但是，它在windows 32应用程序中工作，而不是在控制台32应用程序中工作。并且，您不能在控制台程序集应用程序中使用io.h。不过io.h在windows32汇编程序中可以工作。如果您愿意，我可以向您展示我制作的一个使用io.h并可以工作的windows32汇编程序。我已经做了大约3个这样的工作。。。。谷歌it老兄。@RossRidge买这本书：你需要研究那本书。您需要遵循在windows32汇编程序中使用io.h的编程练习。我告诉你，它是有效的。代码是本书的核心部分，它是有效的。我只是用了控制台应用程序projrct而不是windows应用程序。@MichaelPetch您可能是对的，这里涉及了不同的

io.h

，但OP“GeekDewd”对它在“控制台应用程序”中不起作用提出了截然不同的论点。他之所以没有说他使用了书中的

io.h

，是因为在你发表评论之前，他根本不知道它的存在。

.586
.MODEL FLAT
INCLUDE io.h            ; header file for input/output

.STACK 4096

.DATA
firstInput DWORD    ?                       ; Reserve DWORD sized named memory for 1st input
secondInput DWORD   ?                       ; Reserve DWORD sized named memory for 2nd input
prompt1 BYTE    "Enter first number",  0    ; Prompt user for the first input
prompt2 BYTE    "Enter second number", 0    ; Prompt user for the second input
inputString  BYTE    11 DUP (?)             ; Reserve memory for to store inputs 
                                            ; as integer max integer is 10 digits big
resultLbl BYTE  "GCD", 0                    ; label for dialogBox

resultLb2 BYTE  "The GCD is:"               ; String to be diplayed in the dialogBox
gcdResult BYTE  10 DUP (?), 0               ; to display " The GCD is:" and then gcdInput

.CODE
_MainProc PROC
    ;************************ GET FIRST INPUT ************************
        input   prompt1, inputString, 11    ; read ASCII characters
                                            ; Max length for integer is 10 digits
        atod    inputString                 ; convert to integer
        mov     firstInput, eax             ; store in memory
    ;************************ GET SECOND INPUT ***********************
                                            ; repeat for second number
        input   prompt2, inputString, 11    ; read ASCII characters
        atod    inputString                 ; convert to integer
        mov     secondInput, eax            ; store in memory

        ;******************** THEORY IN PSUDOCODE  *******************
        ; gcdInput is firstInput
        ; remainder is secondInput
        ; loop:
        ; dividend is gcdInput
        ; gcdInput is remainder
        ; 
        ; firstInput and secondInput are reserved memory
        ; dividend is stored in  EAX ---> numerator
        ; gcdInput is stored in  EBX ---> denominator       
        ; remainder is stored in EDX ---> remainder
        ; quotient  is stored in EAX ---> quotient      
        ; EAX/EBX ---> quotient will be stored in EAX
        ;             remainder will be stored in EDX
        ;             based on CPU logic
    ;************************ FIND GCD OF 2 NUMBERS ******************
        mov     EBX, firstInput     ; Copy named memory to EBX
        mov     EDX, secondInput    ; Copy named memory to EDX 

findTheGCD:                 ; DO
        mov     EAX, EBX    ;   Copy gcdInput to EAX. Second pass and so on will
                            ;   copy the denominator to the numerator
        mov     EBX, EDX    ;   Copy remainder to EBX Second pass and so on will
                            ;   copy the remainder to the denominator
        mov     EDX, 0      ;   cdq is for signed, so instead of cdq move 0 to EDX.
                            ;   cdq expands EAX to EDX:EAX filling EDX with the sign bit
                            ;   Since our numbers are positive our sign bit is 0
                            ;   So filling EDX with 0's is the same as cdq would do.
                            ;   We are not using cdq and we are using div because we 
                            ;   are working with unsigned integers. 
                            ;   Clear out previous remainder
        div     EBX         ;   Divide EDX:EAX/EBX (dividend/gcdInput)
                            ;   second pass and so on will divide
                            ;   the denominator by the numerator repeatedly
                            ;   until the remainder is 0
                            ;   Quotient does not matter in this situation
                            ;   After the division process EAX will contain the quotient
                            ;   And EDX will contain the remainder, thus the 0's we filled
                            ;   it with will be overwritten with the remainder
        cmp     EDX, 0      ;   Compare EDX (remainder) to 0
                            ;   Since the 0's were overwritten with the remainder after the division
                            ;   the compare will see the new remainder NOT the 0's that was in it before!!
        jne     findTheGCD  ; UNTIL ( remainder == 0 )

    ;************************ PRINT THE GCD TO SCREEN ******************     
        dtoa    gcdResult, EBX        ; convert to ASCII characters
        output  resultLbl, resultLb2  ; output label and gcd string

        mov     eax, 0  ; exit with return code 0
        ret
_MainProc ENDP
END 

.586
.MODEL FLAT
INCLUDE io.h            ; header file for input/output

.STACK 4096

.DATA
firstInput DWORD    ?                       ; Reserve DWORD sized named memory for 1st input
secondInput DWORD   ?                       ; Reserve DWORD sized named memory for 2nd input
prompt1 BYTE    "Enter first number",  0    ; Prompt user for the first input
prompt2 BYTE    "Enter second number", 0    ; Prompt user for the second input
inputString  BYTE    11 DUP (?)             ; Reserve memory for to store inputs 
                                            ; as integer max integer is 10 digits big
resultLbl BYTE  "GCD", 0                    ; label for dialogBox

resultLb2 BYTE  "The GCD is:"               ; String to be diplayed in the dialogBox
gcdResult BYTE  10 DUP (?), 0               ; to display " The GCD is:" and then gcdInput

.CODE
_MainProc PROC
    ;************************ GET FIRST INPUT ************************
        input   prompt1, inputString, 11    ; read ASCII characters
                                            ; Max length for integer is 10 digits
        atod    inputString                 ; convert to integer
        mov     firstInput, eax             ; store in memory
    ;************************ GET SECOND INPUT ***********************
                                            ; repeat for second number
        input   prompt2, inputString, 11    ; read ASCII characters
        atod    inputString                 ; convert to integer
        mov     secondInput, eax            ; store in memory

        ;******************** THEORY IN PSUDOCODE  *******************
        ; gcdInput is firstInput
        ; remainder is secondInput
        ; loop:
        ; dividend is gcdInput
        ; gcdInput is remainder
        ; 
        ; firstInput and secondInput are reserved memory
        ; dividend is stored in  EAX ---> numerator
        ; gcdInput is stored in  EBX ---> denominator       
        ; remainder is stored in EDX ---> remainder
        ; quotient  is stored in EAX ---> quotient      
        ; EAX/EBX ---> quotient will be stored in EAX
        ;             remainder will be stored in EDX
        ;             based on CPU logic
    ;************************ FIND GCD OF 2 NUMBERS ******************
        mov     EBX, firstInput     ; Copy named memory to EBX
        mov     EDX, secondInput    ; Copy named memory to EDX 

findTheGCD:                 ; DO
        mov     EAX, EBX    ;   Copy gcdInput to EAX. Second pass and so on will
                            ;   copy the denominator to the numerator
        mov     EBX, EDX    ;   Copy remainder to EBX Second pass and so on will
                            ;   copy the remainder to the denominator
        mov     EDX, 0      ;   cdq is for signed, so instead of cdq move 0 to EDX.
                            ;   cdq expands EAX to EDX:EAX filling EDX with the sign bit
                            ;   Since our numbers are positive our sign bit is 0
                            ;   So filling EDX with 0's is the same as cdq would do.
                            ;   We are not using cdq and we are using div because we 
                            ;   are working with unsigned integers. 
                            ;   Clear out previous remainder
        div     EBX         ;   Divide EDX:EAX/EBX (dividend/gcdInput)
                            ;   second pass and so on will divide
                            ;   the denominator by the numerator repeatedly
                            ;   until the remainder is 0
                            ;   Quotient does not matter in this situation
                            ;   After the division process EAX will contain the quotient
                            ;   And EDX will contain the remainder, thus the 0's we filled
                            ;   it with will be overwritten with the remainder
        cmp     EDX, 0      ;   Compare EDX (remainder) to 0
                            ;   Since the 0's were overwritten with the remainder after the division
                            ;   the compare will see the new remainder NOT the 0's that was in it before!!
        jne     findTheGCD  ; UNTIL ( remainder == 0 )

    ;************************ PRINT THE GCD TO SCREEN ******************     
        dtoa    gcdResult, EBX        ; convert to ASCII characters
        output  resultLbl, resultLb2  ; output label and gcd string

        mov     eax, 0  ; exit with return code 0
        ret
_MainProc ENDP
END