mov ah,4ch int 21H
main endp
skiplines proc near
mov bx,ax repeat: cmp bx,0 jz exit
lea dx,crlf mov ah,9 int 21h
dec bx jmp repeat
exit: ret
skiplines endp
s3 ends
end main
19.设有10个学生成绩分别是76, 69，84，73，88，99，63，100和80。试编写一个子程序统计60-69分，70-79分，80-89分，90-99分和100分的人数，并分别放到S6，S7，S8，S9，S10单元中
s1 segment stack
dw 100h dup(?) top label word s1 ends
s2 segment
score dw 76,69,84,90,73,88,99,63,100,80,'$' s6 dw 0,'$' s7 dw 0,'$' s8 dw 0,'$' s9 dw 0,'$' s10 dw 0,'$' s2 ends
s3 segment
assume cs:s3,ds:s2,ss:s3 main proc far
mov ax,s1 mov ss,ax lea sp,top mov ax,s2 mov ds,ax mov si,0
repeat: cmp si,20 jz exit mov ax,score[si] cmp ax,100 jnz l mov bx,s10[0] inc bx mov s10[0],bx jmp l5 ;必须跳转到si加二那里
l: cmp ax,90 jb l2 mov bx,s9[0] inc bx mov s9[0],bx jmp l5
l2: cmp ax,80 jb l3 mov bx,s8[0] inc bx mov s8[0],bx jmp l5
l3: cmp ax,70 jb l4 mov bx,s7[0] inc bx mov s7[0],bx jmp l5
l4: mov bx,s6[0] inc bx mov s6[0],bx
l5: add si,2
jmp repeat
exit: mov ah,4ch int 21h
mainendp s3 ends
end main
20.编写子程序嵌套结构的程序，把整数分别用二进制和八进制显示出来
s1 segment stack
dw 100h dup(?) top label word s1 ends
s2 segment
data dw 16h,32h,'$' ;要显示的两个数字 crlf db 0dh,0ah,24h,'$' s2 ends
s3 segment
assume cs:s3,ds:s2,ss:s3 bando proc far mov ax,s1 mov ss,ax lea sp,top mov ax,s2 mov ds,ax mov si,0
repeat: cmp si,4 jz exit mov bx,data[si] call pairs
call print_crlf add si,2 jmp repeat
exit: mov ah,4ch int 21h
bando endp
pairs proc near mov cx,10h
l: rol bx,1 ;以二进制数输出—— jc print_one mov dl,'0' jmp print_zero
print_one: mov dl,'1'
print_zero: mov ah,2 int 21h loop l mov dl,' ' mov ah,2 int 21h mov dh,0 mov ch,4 mov cl,3 mov dl,bl and dl,7d add dl,30h push dx
rotate: ror bx,cl mov dl,bl and dl,7d add dl,30h push dx dec ch cmp ch,0 jnz rotate ror bx,1 mov dl,bl and dl,1 add dl,30h
;——以二进制数输出 ;以八进制数输出—— mov ah,2 int 21h
print: cmp ch,5 jz atend pop dx mov ah,2 int 21h inc ch jmp print ;——以八进制数输出
atend: ret
pairs endp
print_crlf proc near lea dx,crlf mov ah,9 int 21h ret
print_crlf endp
s3 ends
end bando
21.在D盘根目录建立一个文件abc.txt，第一次向文件写入“123456”六个字符，第二次增加“abcdefg”几个字符
s1 segment stack
dw 100h dup(?) top label word s1 ends
s2 segment fn db 'd:\\abc.txt',0 fh dw ? buff1 db '1','2','3','4','5','6','$' buff2 db 'a','b','c','d','e','f','$' s2 ends
s3 segment
assume cs:s3,ds:s2,ss:s3