请大虾帮忙解释一下！

erwinrommel 2004-07-22 04:57:09

ｍｏｖ　bx,100;
call sound;

sound proc near
push ax ;save registers
push bx
push cx
push dx
push di
mov di,tone ;传递频率
mov al,0b6h ;write timer mode reg.
out 43h,al
mov dx,12h ;timer divisor
mov ax,348ch
div di ;1193100hz/freq 机器频率除以音符频率，人可以听得见
out 42h,al ;write timer cout low byte
mov al,ah
out 42h,al ;write timer cout high byte
in al,61h ;get current port setting
mov ah,al ;and save it in ah
or al,3h ;turn speaker on
out 61h,al
wait2:
mov cx,633
call waitf
dec bx
jnz wait2
mov al,ah
out 61h,al
pop di
pop dx
pop cx
pop bx
pop ax
ret
sound endp
; 等待，延时一点
waitf proc near
push ax
wait1:
in al,61h
and al,10h
cmp al,ah
je wait1
mov ah,al
loop wait1
pop ax
ret
waitf endp
请注释详细一点！分也不多，只给一人，小弟我谢了！

...全文

126 9 打赏收藏转发到动态举报

写回复

用AI写文章

9 条回复

切换为时间正序

请发表友善的回复…

发表回复

snowegale888 2004-07-25

打赏
举报

详细的可以买一本ibm pc汇编的习题包括答案。。绝对好用

snowegale888 2004-07-25

打赏
举报

wait2:
mov cx,633 和waitf的loop wait1对应循环次数
call waitf 调用waitf 应该是关闭发生吧and al,10h
dec bx bx应该是发生次数吧减一
jnz wait2 =0 jmp
mov al,ah 还原端口61初始状态ah 里是备份
out 61h,al 送入61h
pop di 还原寄存器
pop dx
pop cx
pop bx
pop ax
ret
sound endp
; 等待，延时一点
waitf proc near
push ax
wait1:
in al,61h
and al,10h
cmp al,ah
je wait1
mov ah,al
loop wait1
pop ax
ret
waitf endp

SDFDSAC 2004-07-25

打赏
举报

call waitf ;是为了延时,waitf是个延时子程序

erwinrommel 2004-07-23

打赏
举报

wait2:
mov cx,633
call waitf
dec bx
jnz wait2
mov al,ah
out 61h,al
pop di
pop dx
pop cx
pop bx
pop ax
ret
sound endp
; 等待，延时一点
waitf proc near
push ax
wait1:
in al,61h
and al,10h
cmp al,ah
je wait1
mov ah,al
loop wait1
pop ax
ret
waitf endp

这里看不懂

SDFDSAC 2004-07-23