我创建了这个程序来计算毕业所需的学分，从基本用户输入的名称，学位，学位所需学分和当前学分获得。汇编对我来说是新的，到目前为止，我们只学习了I / O和消息框的基础知识，在本例中是“标签”。虽然我只被要求通过消息框输出学位所需的学分我认为探索如何显示用户输入的名称和学位可能会很有趣。我的程序在一个消息框中显示名称和学位，在另一个消息框中显示所需的学分。我的问题是，是否可以将两个框组合成一个显示两组输出的框？提前谢谢。
.586
.MODEL FLAT
INCLUDE io.h ; header file for input/output
.STACK 4096
.DATA
student_name DWORD ?
student_degree DWORD ?
credits_needed DWORD ?
credits_completed DWORD ?
prompt1 BYTE "Please enter your name and degree: ", 0
stringIn BYTE 80 DUP (?)
displayLbl BYTE "Student Name - Student Degree...", 0
stringOut BYTE 80 DUP (?)
prompt2 BYTE "How many credits in your degree field?", 0
prompt3 BYTE "Lastly, how many credits have your earned?", 0
string BYTE 60 DUP (?)
resultLbl BYTE "Credits needed for your chosen degree", 13,10
sum BYTE 11 DUP (?), 0
.CODE
_MainProc PROC
input prompt1, stringIn, 80 ;ask for string of student name and degree
lea eax, stringIn ;source
push eax
lea eax, stringOut
push eax
call strcopy
add esp, 8
input prompt2, string, 60
atod string
mov credits_needed, eax
input prompt3, string, 60
atod string
mov credits_completed, eax
mov eax, credits_needed ; credits needed to EAX
sub eax, credits_completed ; subtract credits earned
dtoa sum, eax ; convert to ASCII characters
output displayLbl, stringOut
output resultLbl, sum ; output label and sum
mov eax, 0 ; exit with return code 0
ret
_MainProc ENDP
strcopy PROC NEAR32
push ebp
mov ebp, esp
push edi
push esi
pushfd
mov edi, [ebp+8]
mov esi, [ebp+12]
cld
whileNoNull:
cmp BYTE PTR [esi], 0
je endWhileNoNull
movsb
jmp whileNoNull
endWhileNoNull:
mov BYTE PTR [edi],0
popfd
pop esi
pop edi
pop ebp
ret
strcopy ENDP
END ; end of source code
引起我注意的是，如果没有看到io.h资源，就不可能这样说：
.586
EXTRN _getInput:NEAR32, _showOutput:NEAR32, atodproc:NEAR32, dtoaproc:NEAR32, wtoaproc:NEAR32, atowproc:NEAR32
dtoa MACRO dest,source ; convert double to ASCII string
push ebx ; save EBX
lea ebx, dest ; destination address
push ebx ; destination parameter
mov ebx, [esp+4] ; in case source was EBX
mov ebx, source ; source value
push ebx ; source parameter
call dtoaproc ; call dtoaproc(source,dest)
add esp, 8 ; remove parameters
pop ebx ; restore EBX
ENDM
atod MACRO source ; convert ASCII string to integer in EAX
lea eax,source ; source address to AX
push eax ; source parameter on stack
call atodproc ; call atodproc(source)
add esp, 4 ; remove parameter
ENDM
wtoa MACRO dest,source ; convert word to ASCII string
push ebx ; save EBX
lea ebx,dest ; destination address
push ebx ; destination parameter
mov ebx, [esp+4] ; in case source was BX
mov bx, source ; source value
push ebx ; source parameter
call wtoaproc ; call dtoaproc(source,dest)
add esp, 8 ; remove parameters
pop ebx ; restore EBX
ENDM
atow MACRO source ; convert ASCII string to integer in AX
lea eax,source ; source address to AX
push eax ; source parameter on stack
call atowproc ; call atodproc(source)
add esp, 4 ; remove parameter
ENDM
output MACRO outLbl, outStr ; display label and string
pushad ; save general registers
cld ; clear DF
lea eax,outStr ; string address
push eax ; string parameter on stack
lea eax,outLbl ; label address
push eax ; string parameter on stack
call _showOutput ; showOutput(outLbl, outStr)
add esp, 8 ; remove parameters
popad ; restore general registers
ENDM
input MACRO inPrompt, inStr, maxLength ; prompt for and input string
pushad ; save general registers
mov ebx, maxLength ; length of input string
push ebx ; length parameter on stack
lea ebx,inStr ; destination address
push ebx ; dest parameter on stack
lea ebx,inPrompt ; prompt address
push ebx ; prompt parameter on stack
call _getInput ; getInput(inPrompt, inStr, maxLength)
add esp, 12 ; remove parameters
popad ; restore general registers
ENDM
.NOLISTMACRO ; suppress macro expansion listings
.LIST ; begin listing
答案 0 :(得分：1)
是否可以将两个框组合成一个显示两组输出的框？
如果您从 stringOut 缓冲区中的 stringIn 中复制输入并同时拥有，则可以轻松地同时显示这两组信息 stringOut 缓冲区附近的resultLbl 和 sum 。
displayLbl BYTE "Student Name - Student Degree...", 0
stringOut BYTE 80 DUP (?), 10
resultLbl BYTE "Credits needed for your chosen degree", 13,10
sum BYTE 11 DUP (?), 0
请注意 stringOut 之后的额外换行符！
input prompt1, stringIn, 80 ;ask for string of student name and degree
lea eax, stringIn ;source
push eax
call szlen ;Gives length in EAX
mov ebx, stringOut+79 ;Position for the terminating zero
sub ebx, eax ; minus the length
push ebx
call strcopy
add esp, 8
mov byte ptr [stringOut+79], 13
... other stuff that doesn't clobber EBX
这里至关重要的是用回车符替换 strcopy 写的终止零，以便所有文本都能无缝连接。
output displayLbl, [ebx]
这是 szlen 宏在 io.h 中定义的方式：
szlen MACRO string ; get string length
lea eax, string ; string address
push eax ; string parameter on stack
call szlenproc ; call szlenproc(string)
ENDM
这就是 szlenproc 过程的样子：
; szlenproc(source)
; Procedure to calculate length of a null-terminated string
; No registers are changed; flags are not affected.
szlenproc PROC NEAR32
push ebp ; save base pointer
mov ebp, esp ; establish stack frame
pushad
pushfd ; save flags
mov esi,[ebp+8] ; source address
mov strAddr, esi
; find string length
mov strLength, 0 ; initialize string length
WhileChar: cmp BYTE PTR [esi], 0 ; character = null?
jz EndWhileChar ; exit if so
inc strLength ; increment character count
inc esi ; point at next character
jmp WhileChar
EndWhileChar:
popfd ; restore flags
popad ; restore registers
pop ebp
mov eax, strLength ; rerun string length in EAX
ret 4 ;exit, discarding parameter
szlenproc ENDP