Question

嘿所以我在运行此代码时收到此错误：

1>------ Build started: Project: Project, Configuration: Debug Win32 ------
1>  Assembling [Inputs]...

1>assign2.asm(32): error A2022: instruction operands must be the same size

1>assign2.asm(33): error A2022: instruction operands must be the same size

1>C:\Program Files (x86)\MSBuild\Microsoft.Cpp\v4.0\BuildCustomizations\masm.targets(49,5):

 error MSB3721: The command "ml.exe /c /nologo /Zi /Fo"Debug\assign2.obj" /Fl"Project.lst" 

/I "c:\Irvine" /W3 /errorReport:prompt  /Taassign2.asm" exited with code 1.

========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

以下代码，当我尝试从usPop中减去ml1337skillz并将结果存储到差异时，会发生这种情况。（请记住，我刚刚开始学习这个领域：））

TITLE Learning      (learning.asm)

INCLUDE Irvine32.inc

secondE = 00000000010001010000000000000000b

.data

myName BYTE "John Smith"

nameLen = ($ - myName)

nationalDebt QWORD 7975482431119.47d

usPop DWORD 313900000d

kindaNegative SDWORD -2147483648d

my1337Sk1LLz WORD 1337h

negUnit SBYTE -1

half BYTE 0.5d

Difference SWORD ?

starField BYTE 2000 DUP("*")

bigLoss SDWORD -50000

.code
main PROC

FillRegs:

    mov eax,usPop           ;store 3139000000d into eax
    sub eax,my1337Sk1LLz    ;subtracts 1337h from usPop in eax
    mov Difference, eax     ;stores eax into Difference

    mov edx,bigLoss         ;stores -50000 into edx

    mov eax,secondE         ;store 0E00 into eax
    mov ah,'A'              ;store string A into ah 
    mov al,'X'              ;store string X into al

    mov ebx,secondE         ;store 0E00 into ebx
    mov bh,'B'              ;store string B into bh
    mov bl,'X'              ;store string X into bl

    call DumpRegs           ;shows Registers

    exit                    ;exits
    main ENDP

END main

Answer 1

这两行是你的问题：

sub eax,my1337Sk1LLz    ;subtracts 1337h from usPop in eax
mov Difference, eax     ;stores eax into Difference

eax是32位，但my1337Sk1LLz和Difference都是16位。

有两种方法可以解决这个问题：

更改my1337Sk1LLz和Difference的尺寸。现在您的类型分别为WORD和SWORD。您可以将这些更改为DWORD和SDWORD，使其成为32位。
零扩展和截断。您需要另一个注册。我将使用edx，因为您似乎没有在那里使用它。首先，您需要签名my1337Sk1LLz：
```
movzx edx, my1337Sk1LLz  ; move, zero-extended, my1337Sk1LLz into EDX
```
然后你可以做减法：
```
sub eax, edx  ; they're the same size now so we can do this
```
然后，您可以将eax的低位词存储到Difference中，弃用高位词：
```
mov Difference, ax
```

错误A2022：指令操作数必须大小相同

1 个答案: