爲什麼在爲第二個矩陣進行用戶輸入時，rd_next代碼永遠不會結束？

Question

下面的代碼應該添加兩個一維矩陣並顯示總和。程序獲取第二個矩陣的輸入時出現問題：rd_next循環永遠不會結束。然而，它需要第一個矩陣的輸入很好。

data_seg segment 
mat1 dw 3 dup(?) 
mat2 dw 3 dup(?) 
n db 3 
ten dw 10 
counter db ? 
string db 10 dup(?) 
msg1 db 10,13,"Enter first matrix: ","$" 
msg2 db 10,13,"Enter second matrix: ","$" 
msg3 db 10,13,"Enter a number: ","$" 
data_seg ends 

code_seg segment 
assume cs:code_seg,ds:data_seg 

print_string proc  
pop si  
pop dx 
mov ah,9   
int 21h  
push si  
ret   
print_string endp 

read_char proc   
pop di 
mov ah,1   
int 21h  
mov ah,0 
push ax  
push di  
ret 
read_char endp 

read_number proc 
pop si  
mov bx,0  
mov dx,0 

next_digit: 
call read_char 
pop ax   
cmp al,0Dh  
je done 
sub al,30h  
mov cl,al  
mov ch,0   
mov ax,bx  
mul ten  
add ax,cx 
mov bx,ax  
jmp next_digit 

done: push bx   
push si 
ret   
read_number endp 

print_number proc  
pop si  
pop ax 
mov bx,0   
mov dx,0 
repeat1:   
mov cx,0   
mov dx,0   
div ten  
push dx  
inc counter  
cmp ax,0   
jne repeat1  

print_digit: 
pop dx  
add dl,30h  
mov ah,2   
int 21h  
dec counter  
jnz print_digit 
push si  
ret   
print_number endp 

start: 
mov ax,data_seg 
mov ds,ax 

mov al,n    
mov counter,al    ; initialize counter variable 
mov bp,offset mat1   ; initialize pointer to first matrix 

push offset msg1   ; prompt user to enter first matrix 
call print_string 

rd_next: 
push offset msg3   ; prompt user for next number in matrix 
call print_string   
call read_number   ; call the read_number procedure. 
pop dx   
mov [bp],dx 
add bp,2   
dec counter 
jnz rd_next     ; loop back to read the next number. 

mov counter,al    ; reset counter variable 
mov bp,offset mat2   ; initialize pointer to second matrix 

push offset msg2   ; prompt user to enter second matrix 
call print_string 
jmp rd_next 

mov al,n 
mov si,offset mat1 
mov di,offset mat2 

matrixsum: 
mov bx,[si] 
mov cx,[di] 
add bx,cx 
push bx 
call print_number 
inc si 
inc di 
dec al 
jnz matrixsum 

code_seg ends 
end start 

我沒有看到任何代碼錯誤。我試圖創建一個單獨的rd_next2循環來獲取第二個矩陣的輸入，但它不起作用。

Answer 1

哦，它會結束所有權利，只需要一些時間，因爲你忘了重新初始化counter。移動初始化代碼rd_next內，例如：

mov bp,offset mat1   ; initialize pointer to first matrix 

    push offset msg1   ; prompt user to enter first matrix 
    call print_string 

rd_next: 
    mov al,n    
    mov counter,al    ; initialize counter variable 

學習如何使用調試器，以便您可以單步執行你的代碼，看看它爲什麼做它是什麼。 PS：你有一種巧妙的方式來返回一個函數的結果，但請不要這樣做：D只需使用一個寄存器作爲其他人。