以下程序的目标是接受最多10个带符号的8字节浮点数 作为用户输入的-100≤X≤100范围内的数字并将它们存储到数组中。使用ReadFloat Irvine方法接收用户输入。如果输入该范围之外的数字，则子程序应该停止执行并通过eax返回当前数组中的值的数量。这只是用来描述程序应该做什么的一些上下文。我对此代码的问题是它在接受第一个值后没有正确循环。我进行了设置，以便检查输入数字是否在L1中高于或等于-100，然后在L2中低于或等于100。如果数字超出该范围，子程序应该停止执行，但如果它在该范围内，它应该进入L3＆amp; R1。在L3和R1中，数字被放入SFPArray中的索引中，如果数组中的值少于10，则程序应无条件地跳回L1以进行进一步的迭代。 R1中的JMP命令是问题所在。在输入单个数字后，当前状态的子程序将停止执行，我无法弄清楚原因。有人可以提供帮助吗？
INCLUDE c:\irvine\irvine32.inc
INCLUDELIB c:\irvine\irvine32.lib
INCLUDELIB c:\masm32\lib\user32.lib
INCLUDELIB c:\masm32\lib\kernel32.lib
.data
theSFPArray REAL8 10 dup(?) ;an array that can store up to 10 signed floating point numbers
tempStoreFP REAL8 ? ;this variable will temporarily store the FP number acquired from user input, and then push it onto the stack
lengthOfSFPArray DWORD ? ;this variable will store the length of theSFPArray. This value will be used to determine if requestSignedFloats should stop looping.
inputLoopCounter DWORD -1 ;used to determine when the requestSignedFloats subroutine should stop accepting input.
prompt BYTE "Please enter a value: ",0
.CODE
main PROC
call requestSignedFloats
exit
main ENDP
requestSignedFloats PROC
finit ;initializes floating point unit
push edx ;pushes the original value of edx onto the stack. This will be popped when the subroutine ends.
mov edx, OFFSET theSFPArray ;moves the offset of theSFPArray into edx so that values can be placed into it.
push edx ;pushes edx onto the stack while it contains the offset of the SFPArray for later usage.
mov eax,100
push eax
fild dword ptr [esp] ;get the 100 from memory and throw it onto the FPU, into ST(0)
fchs ;changes the 100 in ST(0) into -100
pop eax
L1:
mov edx,OFFSET prompt
call WriteString ;displays the String within the prompt variable on the screen.
call ReadFloat ;requests a float as input from the user and stores it at the top of the floating point stack, aka ST(0).
fcom ;compares the value in ST(1) to the value in ST(0).
jae L2
pop edx ;this line and the two lines below it will execute if the comparison dictates that ST(1) is below the value in ST(0). This should cause the subroutine to end.
pop edx ;a second pop of edx is necessary to restore edx to its original value since two alterations of edx were placed onto the stack at the beginning of the subroutine.
mov lengthOfSFPArray,LENGTHOF theSFPArray ;Moves the current number of values stored in theSFPArray into the lengthOfSFPArray variable.
mov eax,lengthOfSFPArray ;Returns in eax,the number of values in the array, as specified by the requirements
ret
L2:
fstp tempStoreFP ;pops the user input value off of the stack temporarily so that fchs can be used to change the sign of the value in ST(0)
fchs ;changes the -100 in ST(0) into a positive 100.
fld tempStoreFP ;pushes tempStoreFP back onto the stack so that its value is now in ST(1)
fcom
jbe L3
pop edx ;this line and the two lines below it will execute if the comparison dictates that ST(1) is below the value in ST(0). This should cause the subroutine to end.
pop edx ;a second pop of edx is necessary to restore edx to its original value since two alterations of edx were placed onto the stack at the beginning of the subroutine.
mov lengthOfSFPArray,LENGTHOF theSFPArray ;Moves the current number of values stored in theSFPArray into the lengthOfSFPArray variable.
mov eax,lengthOfSFPArray ;Returns in eax,the number of values in the array, as specified by the requirements
ret
L3:
pop edx ;this is done to pop the offset of theSFPArray off of the stack and back into edx since at this point edx still stores the "prompt".
inc inputLoopCounter ;increments inputLoopCounter so that its value is equal to the index that the number input by the user will be stored in.
mov ecx,inputLoopCounter ;uses inputLoopCounter to determine how many times the loop will execute.
R1:
inc edx ;increments edx an amount of times equivalent to the value stored in inputLoopCounter.
loop R1
fstp qword ptr [edx] ;takes the value at the top of the stack and stores it as a REAL8 at the address specified by edx (aka its array index)
mov lengthOfSFPArray,LENGTHOF theSFPArray ;Moves the current number of values stored in theSFPArray into the lengthOfSFPArray variable.
fchs ;changes the 100 in ST(0) to a -100 in preparation for the next iteration of the subroutine.
cmp inputLoopCounter,10
je L4
jmp L1 ;An unconditional jump to L1 that causes this subroutine to execute repeatedly. The line above this one prevents it from being an infinite loop.
L4:
mov eax,lengthOfSFPArray ;Returns in eax,the number of values in the array, as specified by the requirements
pop edx ;if the program makes it to this point, the offset of the array would have been popped off of the stack, meaning the original value of edx is the only thing
;remaining on the stack, so only one pop is necessary
ret
requestSignedFloats ENDP
答案 0 :(得分：5)
在products部分中，您可以像这样定义.data
lengthOfSFPArray lengthOfSFPArray DWORD ? ;this variable will store the length of theSFPArray. This value will be used to determine if requestSignedFloats should stop looping.
表示初始值未定义，因此介于0和2 ^ 32-1之间。
在 L1 中，您使用
检索?值
undefined因此mov eax,lengthOfSFPArray ;Returns in eax,the number of values in the array, as specified by the requirements
将是EAX或初始化时undefined的值{}。你在 L2 中重复一遍。
在 R1 中，您可以使用
设置lengthOfSFPArray
lengthOfSFPArray到mov lengthOfSFPArray,LENGTHOF theSFPArray
的LENGTHOF，在theSFPArray部分中定义为
data根据定义，theSFPArray REAL8 10 dup(?)
：theSFPArray中的元素数量。
之后，将10的值与10的值进行比较，该值始终为LENGTHOF(theSFPArray) = 10：
TRUEL4：是您的退出标签，因此整个程序只执行一次。