汇编里头，输入6个元素的数组的值，求和！----累加器

这篇博客展示了如何使用汇编语言编程，输入6个整数存储到数组中，然后通过累加器计算这些元素的总和，并将结果显示出来。主要涉及汇编里的数据段、栈段、代码段的定义，以及输入、输出和循环控制等操作。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

data segment
ary dw 10 dup(?)
sum dw ?
temp db 10 dup(?)
data ends

stacker segment stack
    dw 20 dup(?)
top label word
stacker ends

code segment
    assume cs:code,ds:data,ss:stacker

main proc near

    mov ax,data
    mov ds,ax
    mov ax,stacker
    mov ss,ax
    lea sp,top

    mov cx,6
    mov si,0
L1:
    xor bx,bx
    call near ptr input
    call near ptr line
    mov ary[si],bx
    add sum,bx
    add si,2
    loop L1


    mov ax,sum
    mov di,0
    mov cl,10
L2:
    div cl
    mov temp[di],ah
    cmp al,0
    je L3
    add di,2
    xor ah,ah
    jmp L2

L3:
    call near ptr print

    mov ah,4ch
    int 21h
main endp

input proc near
    push ax
    push dx
    push di

    mov di,10d

L4:
    mov ah,01h
    int 21h
    cmp al,0dh
    je L5
    cmp al,'0'
    jb L4
    cmp al,'9'
    ja L4
    sub al,30h
    xor ah,ah
    xchg ax,bx
    mul di
    xchg ax,bx
    add bx,ax
    jmp L4

L5:
    ;call near ptr line
    ;mov dx,bx
    ;add dl,30h
    ;mov ah,02h
    ;int 21h

    pop di
    pop dx
    pop ax
    ret
input endp

print proc near
    push di
    push dx
    push ax

L6:
    mov dl,temp[di]
    add dl,30h
    mov ah,02h
    int 21h
    cmp di,0
    je L7
    sub di,2
    jmp L6

L7:
    pop ax
    pop dx
    pop di
    ret
print endp

line proc near
push ax
push dx
mov dl,0dh
mov ah,02h
int 21h
mov dl,0ah
mov ah,02h
int 21h
pop dx
pop ax
ret
line endp

code ends
end main