我的代码中存在问题。
我正在尝试将DI寄存器的值移动到名为 var 的变量中。
DI寄存器中的值可以是0,1,2或3（随机选择），但是当我将其移动到 var 时，值将更改为514（I使用调试器中的watch功能来查看）
问题出现在“游戏”标签下的第440行。
IDEAL
MODEL small
STACK 100h
DATASEG
x dw 80
y dw 64
green db 2
red db 4
yellow db 14
blue db 1
white db 15
arr dw 11 dup(4)
Clock equ es:6Ch
hold db 0
key db 0
marker dw 0
var dw ?
CODESEG
proc YellowP
mov[x], 80
mov[y], 64
yLoop:
mov bh,0h
mov cx,[x]
mov dx,[y]
mov al,[yellow]
mov ah,0ch
int 10h
inc [x]
cmp cx, 112
jl yLoop
ySquare:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 96
jl yLoop
ret
endp YellowP
proc BlueP
mov[x], 113
mov[y], 64
bLoop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [blue]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl bLoop
bSquare:
mov bh, 0h
inc [y]
mov [x], 113
cmp dx, 96
jl bLoop
ret
endp BlueP
proc GreenP
mov[x],80
mov[y],32
gLoop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [green]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 112
jl gLoop
gSquare:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 64
jl gLoop
ret
endp GreenP
proc RedP
mov[x],112
mov[y],32
rLoop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [red]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl rLoop
rSquare:
mov bh, 0h
inc [y]
mov [x], 112
cmp dx, 64
jl rLoop
ret
endp RedP
;===============================================================
;===============================================================
proc Random
RND:
mov ax, 40h
mov es, ax
mov ax, [es:6Ch]
and al, 00000011b
xor ah, ah
mov di, ax
cmp al, 0
je GreenPress
cmp al, 1
je RedPress
cmp al, 2
je YellowPress
jg BlueMid
GreenPress:
mov[x],80
mov[y],32
g2Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 112
jl g2Loop
g2Square:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 64
jl g2Loop
ret
RedPress:
mov[x],112
mov[y],32
r2Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl r2Loop
r2Square:
mov bh, 0h
inc [y]
mov [x], 112
cmp dx, 64
jl r2Loop
ret
BlueMid:
jmp BluePress
YellowPress:
mov[x], 80
mov[y], 64
y2Loop:
mov bh,0h
mov cx,[x]
mov dx,[y]
mov al,[white]
mov ah,0ch
int 10h
inc [x]
cmp cx, 112
jl y2Loop
y2Square:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 96
jl y2Loop
ret
BluePress:
mov[x], 113
mov[y], 64
b2Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl b2Loop
b2Square:
mov bh, 0h
inc [y]
mov [x], 113
cmp dx, 96
jl b2Loop
ret
endp Random
proc second
mov ax, 40h
mov es, ax
mov ax, [Clock]
FirstTick:
cmp ax, [Clock]
je FirstTick
mov cx, 9
DelayLoop:
mov ax, [Clock]
Tick:
cmp ax, [Clock]
je Tick
loop DelayLoop
ret
endp second
;===============================================================
;===============================================================
proc Gpress
mov[x],80
mov[y],32
g3Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 112
jl g3Loop
g3Square:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 64
jl g3Loop
ret
endp Gpress
proc Rpress
mov[x],112
mov[y],32
r3Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl r3Loop
r3Square:
mov bh, 0h
inc [y]
mov [x], 112
cmp dx, 64
jl r3Loop
ret
endp Rpress
proc Ypress
mov[x], 80
mov[y], 64
y3Loop:
mov bh,0h
mov cx,[x]
mov dx,[y]
mov al,[white]
mov ah,0ch
int 10h
inc [x]
cmp cx, 112
jl y3Loop
y3Square:
mov bh, 0h
inc [y]
mov [x], 80
cmp dx, 96
jl y3Loop
ret
endp Ypress
proc Bpress
mov[x], 113
mov[y], 64
b3Loop:
mov bh, 0h
mov cx, [x]
mov dx, [y]
mov al, [white]
mov ah, 0ch
int 10h
inc [x]
cmp cx, 144
jl b3Loop
b3Square:
mov bh, 0h
inc [y]
mov [x], 113
cmp dx, 96
jl b3Loop
ret
endp Bpress
proc RNDsquare
call second
call random
call second
cmp di, 0
je aG
cmp di, 1
je aR
cmp di, 2
je aY
jg aB
aG:
call GreenP
jmp endproc
aR:
call RedP
jmp endproc
aY:
call YellowP
jmp endproc
aB:
call BlueP
jmp endproc
endproc:
ret
endp RNDsquare
proc CheckArr
mov si, 0
check:
cmp [arr + si], 0
je tempo1
cmp [arr + si], 1
je tempo2
cmp [arr + si], 2
je tempo3
cmp [arr + si], 3
je tempo4
jg tempo
tempo1:
call second
call Gpress
call second
call GreenP
inc si
jmp check
tempo2:
call second
call Rpress
call second
call RedP
inc si
jmp check
tempo3:
call second
call Ypress
call second
call YellowP
inc si
jmp check
tempo4:
call second
call Bpress
call second
call BlueP
inc si
jmp check
tempo:
mov si, 0
ret
endp CheckArr
;===============================================================
;===============================================================
start:
mov ax, @data
mov ds, ax
; Graphic mode
mov ax, 13h
int 10h
call YellowP
call GreenP
call BlueP
call RedP
call RNDsquare
mov [arr + 0], di
mov ah, 1
int 21h
mov [key], al
cmp di, 0
je temp1
cmp di, 1
je temp2
cmp di, 2
je temp3
jg temp4
temp1:
cmp [key], 113
jne nextmid
call Gpress
call second
call GreenP
je prog
temp2:
cmp [key], 119
jne nextmid
call Rpress
call second
call RedP
je prog
temp3:
cmp [key], 97
jne nextmid
call Ypress
call second
call YellowP
je prog
temp4:
cmp [key], 115
jne nextmid
call Bpress
call second
call BlueP
je prog
;=========================================after round 1
nextmid:
jmp next
prog:
call CheckArr
cmp [arr + 12], 4
jne nextmid2
inc [marker]
mov si, [marker]
call RNDsquare
mov [arr + si], di
mov si, 0
game:
mov di, [arr + si]
mov [var], di
cmp [var], 0
je game0
cmp [var], 1
je game1
cmp [var], 2
je game2
cmp [var], 3
je game3
jmp prog
game0:
mov ah, 1
int 21h
cmp al, 113
jne next
call Gpress
call second
call GreenP
inc si
jmp game
game1:
mov ah, 1
int 21h
cmp al, 119
jne next
call Rpress
call second
call RedP
inc si
jmp game
nextmid2:
jmp next
game2:
mov ah, 1
int 21h
cmp al, 97
jne next
call Ypress
call second
call YellowP
inc si
jmp game
game3:
mov ah, 1
int 21h
cmp al, 115
jne next
call Bpress
call second
call BlueP
inc si
jmp game
next:
; Wait for key press
mov ah,00h
int 16h
; Return to text mode
mov ah, 0
mov al, 2
int 10h
exit:
mov ax, 4c00h
int 21h
END start
答案 0 :(得分：1)
arr dw 11 dup(4)
您的程序的基本问题是您已将数组 arr 定义为 word ，但您反复遍历数组，就好像它包含字节一样即可。
重新定义阵列将是最好的，并将避免大量其他修改。所以我建议一个简单的解决方案，以便能够继续使用DI寄存器：
arr db 11 dup(4)
...
game:
mov di, [arr + si]
and di, 255 <<<< removes the garbage high byte!
mov [var], di
这样mov di, [arr + si]指令遗憾地读取了一个垃圾字节，但是下面的and di, 255只是清除它。简单而且省力。
您的程序中有一个位于数组中的位置。这里只是编写DI寄存器不再是一个选项。您将不得不使用字节大小的寄存器：
mov ax, di ;Temporarily copy to splittable register
mov [arr + si], al ;Only write the low byte
mov si, 0
game:
我认为阵列不够大。在一行中你写了cmp [arr + 12], 4。这没有意义！对于字大小的数组，它指向中间的某个位置，对于字节大小的数组，它将指向数组
您使用值 4 初始化数组这一事实意味着您需要检查数组中的最后一个元素。这是在11个元素字节大小的数组中的偏移量10。
arr db 11 dup(4)
...
prog:
call CheckArr
cmp [arr + 10], 4 <<< checks last element!
答案 1 :(得分：0)
快速查看game:
mov di, [arr + si]
mov [var], di
di是16b寄存器，所以你从[arr + si]中取出 word （16b），你准备了一些随机0-3 字节（8b）我猜？
514为0x0202，实际上它看起来像是两个字节2。
因此，如果var为16b，而arr为8b源代码，则可以通过各种方式进行修复，然后使用：
movzx di, byte [arr + si] ; fetch only 8b from []
; and extend them to 16b with zeroes (zx = zero extended)
; (similar instruction movsx extends with sign bit)
mov [var], di ; store the converted 0-3 16b value to var
编辑：
虽然你的arr dw 11 dup(4)定义表明arr意在用作16b数组，但真正的问题可能在你的随机0-3生成器中，存储字节而不是字。然后在game标签下不需要修复，但在其他地方不需要（我没有费心去搜索arr生成器）。