操作码和操作数的组合无效

Question

SEGMENT .data
    print       db      "%d %d %d %d This is a test of printf", 10, 0
    rowm        dw      160     ;row multiplier
    iterations  db      80      ;number of columns to set

SEGMENT .bss
    offs        resd    1       ;offset 

SEGMENT .text
    attribute   equ     47h ;color attribute

    global _VText
    global _VPage
    extern _vm_buffer
    extern _printf


_VText:

    push ebp
    mov ebp, esp
    push edi
    push esi
    push eax
    push es
    push ecx
    push edx

    mov esi, [ebp+8]        ;message 
    mov es, [_vm_buffer]
    mov dword [offs], 0

    mov ax, [ebp+12]            ;row
    mul dword[rowm]         ;multiply row by 160, result stored in dx:ax
    add [offs], dx          ;add dx:ax to offset
    shl dword [offs], 16
    add [offs], ax

    mov ax, [ebp+16]            ;column
    shl ax, 1               ;multiply column by 2
    add [offs], ax          ;add ax to offset

    mov ax, [ebp+24]            ;page
    shl ax, 12              ;multiply page by 2^12 (4096)
    add [offs], ax          ;add ax to offset   

    mov edi, offs           ;set offset
    mov ah, [ebp+20]        ;attribute

    sub byte[iterations], [ebp+16]   ;so that we don't write too many columns
    mov ecx, iterations

next_char:
    lodsb                   ;get the input string type
    cmp al, 00h             ;check for null character
    je null_ch              ;if null, then quit (null character indicates end of the string)
    stosw                   ;store ax to video memory
    loop next_char          ;will loop 80 times

null_ch:
    pop edx
    pop ecx
    pop es
    pop eax
    pop esi
    pop edi
    pop ebp
    ret


_VPage:



    ret

我之前研究过这个错误，它说添加括号，我这样做了，但它没有修复。

请帮忙。

Answer 1

这是哪种架构，哪种汇编器？对我来说，看起来像 Intel/NASM-ish 语法中的 i386（但这只是一个小片段）。错误出现在哪一行代码上？无论如何你都不能这样做：

sub byte[iterations], [ebp+16]

你不能直接从内存到内存进行减法。您必须通过中间寄存器，例如：

mov eax, [ebp+16]
sub byte[iterations], al

但是您的错误也可能指的是另一行。