Assembly 汇编新手请求帮助：有符号整数除法+&引用；截断为整数；
几天前我问了一个关于这个代码的问题，所以你们中的一些人可能会觉得它很熟悉。对于那些不熟悉它的人来说，这段代码应该做的是向用户请求25个带符号的整数，并将它们存储到一个数组中；这是通过RequestSignedits子例程完成的（我很确定这部分工作正常）。然后，一个“calcMean”子例程将数组中的所有值相加，除以数组中的元素数，然后“将结果截断为整数”。这就是我被困的地方。我试图编写一个calcMean子例程来完成我刚才描述的工作，但似乎无法正确地完成除法。最重要的是，我甚至不确定我当前在calcMean子例程中的内容是否能正常工作。有人能提供帮助吗Assembly 汇编新手请求帮助：有符号整数除法+&引用；截断为整数；,assembly,x86,masm32,irvine32,Assembly,X86,Masm32,Irvine32,几天前我问了一个关于这个代码的问题，所以你们中的一些人可能会觉得它很熟悉。对于那些不熟悉它的人来说，这段代码应该做的是向用户请求25个带符号的整数，并将它们存储到一个数组中；这是通过RequestSignedits子例程完成的（我很确定这部分工作正常）。然后，一个“calcMean”子例程将数组中的所有值相加，除以数组中的元素数，然后“将结果截断为整数”。这就是我被困的地方。我试图编写一个calcMean子例程来完成我刚才描述的工作，但似乎无法正确地完成除法。最重要的是，我甚至不确定我当前在ca
INCLUDE c:\irvine\irvine32.inc
INCLUDELIB c:\irvine\irvine32.lib
INCLUDELIB c:\masm32\lib\user32.lib
INCLUDELIB c:\masm32\lib\kernel32.lib
.data
theSINTArray BYTE 25 dup(?)
lengthOfArray BYTE ?
indexCounter BYTE 0
prompt BYTE "Please enter a value: ",0
.CODE
main PROC
call requestSignedInts
call calculateMean
exit
main ENDP
requestSignedInts PROC
mov edx, offset theSINTArray
Next:
push edx
mov edx,OFFSET prompt
call WriteString
call ReadInt
pop edx
mov [edx], al
inc edx
cmp edx, offset theSINTArray + 25
jb Next
ret
requestSignedInts ENDP
calculateMean PROC
push ecx
mov ecx,lengthOfArray - theSINTArray ; Determine array length
xor eax, eax ; Clear EAX
mov esi, OFFSET theSINTArray ; Starting point for index into array
calcMean:
movsx edx, byte ptr[esi] ; Sign extended move a byte into EDX
add eax, edx ; Accumulate in EAX
inc esi ; Increment source pointer to the next element
loop calcMean ; or cmp esi,endOfArray / jb, then you wouldn't need to touch ecx
mov ecx,lengthOfArray - theSINTArray ; Determine array length
cdq ; sign-extend eax into edx:eax
idiv ecx ; Divide edx:eax by ecx
; eax now contains the integer and edx contains
; the remainder.
pop ecx
ret
calculateMean ENDP
END main
在我看来，您可能缺少注释中显示的
idivcalculateMean PROC
push ecx
mov ecx,lengthOfArray - theSINTArray ; Determine array length
xor eax, eax ; Clear EAX
mov esi, theSINTArray ; Starting point for index into array
calcMean:
movsx edx, byte ptr[esi] ; Sign extended move a byte into EDX
add eax, edx ; Accumulate in EAX
inc esi ; Increment source pointer to the next element
loop calcMean ; or cmp esi,endOfArray / jb, then you wouldn't need to touch ecx
mov ecx,lengthOfArray - theSINTArray ; Determine array length
cdq ; sign-extend eax into edx:eax
idiv ecx ; Divide edx:eax by ecx
; eax now contains the integer and edx contains
; the remainder.
pop ecx
ret
calculateMean ENDP
您已经知道列表的长度，只需清除
edxecxmovsxedx.data
SBYTEArray BYTE 25 dup(?) ; SINT = signed int, which makes most x86 C programmers think 32bit
; lengthOfArray BYTE ? ; totally bogus: named wrong, and there doesn't need to be any storage there
; SBYTEArray_len equ $ - SBYTEArray ; that's NASM syntax and IDK the MASM syntax
SBYTEArray_end: ; SBYTEArray_end and prompt have the same address. That's fine.
prompt BYTE "Please enter a value: ",0 ; If MASM/Windows has a .rodata section, you should put constant data there. It goes in the text section, along with code, because each instance of your process doesn't need a private copy.
.CODE
main PROC
call requestSignedInts
call calculateMean
exit
main ENDP
requestSignedInts PROC
; indent code one level deeper than labels / directives
push esi ; save a call-preserved reg
mov esi, offset SBYTEArray
Next:
mov edx,OFFSET prompt
call WriteString
call ReadInt
mov [esi], al
inc esi
; cmp edx, offset SBYTEArray + 25 ; hard-coding the size defeats the purpose of defining lengthOfArray
cmp esi, offset SBYTEArray_end
jb Next
pop esi ; note that the push/pop are outside the loop
ret
requestSignedInts ENDP
calculateMean PROC
; push ecx ; The normal function-call ABIs allow clobbering ecx, and I don't see any reason to make this function go beyond the ABI requirements (although that is an option in asm)
; push esi ; esi *is* call-preserved in the standard 32bit ABIs. But by changing our function to use fewer registers, we can avoid the save/restore
xor eax, eax ; start with sum=0
mov ecx, offset SBYTEArray
calcMean:
movsx edx, byte ptr[ecx]
add eax, edx
inc ecx
cmp ecx, offset SBYTEArray_end
jb calcMean ; loop while our pointer is below the end pointer
mov ecx, SBYTEArray_end - SBYTEArray ; Determine array length. Does this need OFFSET?
cdq
idiv ecx
; pop esi ; we ended up not needing it
ret
calculateMean ENDP
END Main
特别是当您可以通过使用其他内容来保存寄存器时。除法已经截断为整数，所以什么也不做。顺便说一句，记住edx在除法中的作用。“idiv”指令给出整数结果。正如@harold所说，什么都不做，结果已经在EAX（）中被截断了。@harold如果edx有一个“角色”，很遗憾，我不知道它是什么。我假设这个“角色”应该是常识，但作为一个语言新手，我还没有掌握所有的基本知识。edx的作用到底是什么？在链接中搜索“idiv”，edx是除法的剩余部分，EAX是商，你只需要商，对吗？@JoseManuelAbarcaRodríguez我按照你的建议做了，如果我理解正确，我在代码中的一行：“idiv EAX”，将edx:EAX除以EAX。如果这是正确的，我还有两个问题。1.我应该使用不同的寄存器来存储数组中所有元素的总和吗？2.如何将总和的最高有效四个字节放入edx，最低有效四个字节放入eax？如果
eaxcdqeaxedx:eaxdivcdqidivmovsx[edx+indexCounter]cdqidiv