简单汇编程序设计

做的匆忙，细节还要你自己修改一下：）
第一题
datarea segment

arr db 1，-2，3，0，60，-30，45，52，90，70
len dw 10 ;length of arr

positive db 10 dup(?) ;>0

c1 dw ?

negative db 10 dup(?) ;<0

c2 dw ?
c3 dw ?

datarea ends

coderea segment

main proc far

assume cs:coderea,ds:datarea

start:

push ds

sub ax,ax

push ax

mov ax,datarea

mov ds,ax

mov si,0 ;index of arr

mov di,0 ;index of positive

mov bx,0 ;index of negative

repeat:

cmp si,len

jge exit

cmp arr[si],0

jl movetonegative

jg movetopositive

inc si

jmp repeat

movetonegative:

mov al,arr[si]

mov negative[bx],al

inc bx

inc si

jmp repeat

movetopositive:

mov al,arr[si]

mov positive[di],al

inc di

inc si

jmp repeat

exit:

add di,1

add bx,1

mov c1,di

mov c2,bx

add c1,30h

mov dl,byte ptr c1 ;output the length of positive

mov ah,2

int 21h

add c2,30h

mov dl,byte ptr c2 ;output the length of negative

mov ah,2

int 21h

mov dl,byte ptr len

add dl,30h

mov ah,2

int 21h

ret

main endp

coderea ends

end

第二题
data segment
ary dw 1，-2，3，0，60，-30，45，52，90，70
conut dw 10
s1 dw ?
data ends

stack segment
dw 100 dupOP(?)
tos label word
stack ends

code1 segment
main proc far
assume cs:code1, ds:data, ss:stack
start:
mov ax, stack
mov ss, ax
mov sp, offset tos
push ds
sub ax, ax
push ax
mov ax, data
mov ds, ax
mov bx, offset ary
push bx
mov bx, offset count
push bx
mov bx, offset s1
push bx
call far ptr proadd
ret
main endp
code1 ends

code2 segment
assume cs:code2
proadd proc far
push bp
mov bp, sp

push ax
push cx
push si
push di

mov si, [bp+0ah]
mov di, [bp+8]
mov cx, [di]
mov di, [bp+6]

code2 ends
end start

xor ax, ax
next:
add ax, [si]
add si, 2
loop next
mov [di], ax

pop di
pop si
pop cx
pop ax

pop bp

ret 6
proadd endp