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我正在开发一个在 PC 扬声器上播放音乐的汇编程序 (8086)。

一切正常，但我有一个问题。程序在第 78 个音符处进入无休止的睡眠（扬声器打开），无论它是什么音符。

我正在使用86h function of 15 interrupt.

那么为什么会出现这种无休止的睡眠，以及如何解决呢？

这是代码（带有真人快打主题）：

;constants
TB equ 38636; 1.19MHz/30Hz
TC equ 34546; 1.19MHz/33Hz
TD equ 30811; 1.19MHz/37Hz
TE equ 27805; 1.19MHz/41Hz
TF equ 25909; 1.19MHz/44Hz
TG equ 23265; 1.19MHz/49Hz
TA equ 20727; 1.19MHz/55Hz
TH equ 18387; 1.19MHz/62Hz
TP equ 1;pause
;K end of melody
strophe1n equ 'ACADAEDCCECGCECGGHGCGDCFFAFCFCH'
strophe1o equ '3434344444444443333434433334343'
strophe1t equ '2222222222222222222222222222222'
strophe2n equ 'APAPAPAPGCAPAPAPAPGEAPAPAPAPGCAPAPAPAPAPAPA'
strophe2o equ '4040404045404040404440404040454040404040404'
strophe2t equ '2424242422242424242224242424222424284844484'
strophe3n equ 'AEACABACABGAEACABACABGAEACABACABGAEACGPGGPGAAPAA'
strophe3o equ '454545454544545454545445454545454454540440444044'
strophe3t equ '424242424424242424244242424242442424248488448848'

Progr           segment
                assume  cs:Progr, ds:data, ss:stacky
interval:; waits DX:CX microseconds
        mov ah,86h;
        int 15h
        ret
        ;here come notes every note set up its time and sleep
note:;0,5 sec
        mov cx,7;
        mov dx,41248;
        call interval
        ret

halfnote:;0,25sec
        mov cx,3;
        mov dx,53280;
        call interval
        ret

quarternote:;0,125sec
        mov cx,1;
        mov dx,58982;
        call interval
        ret

eighthnote:
        mov cx,0;
        mov dx,62455;
        call interval
        ret

turnon:;sets tone and turn on speaker
        ;setting up tone
        mov ax,tone
        mov dx,42h
        out dx,al
        mov al,ah
        out dx,al

        ;turning speaker on
        mov dx,61h
        in al,dx;
        or al,00000011B;
        out dx,al;
        ret

turnoff:;turning speaker off
        mov dx,61h
        in al,dx;
        and al,11111100B;
        out dx,al;
        ret

play:
;simple switch for times
        call turnon
        cmp time,1
        je whole
        cmp time,2
        je half
        cmp time,4
        je quarter
        cmp time,8
        je eighth

        whole: call note;sleep for note time (while speaker is on) and then shuts up the speaker
        jmp endplay
        half: call halfnote
        jmp endplay
        quarter: call quarternote
        jmp endplay
        eighth: call eighthnote
        jmp endplay
        endplay:
        call turnoff; turning speaker off
        ret

exit:
        mov ah,4ch
        mov al,00h
        int 21h;
;START--------------------------------------------------------------------
start: mov ax,data ;some start up
mov ds,ax
mov ax,stacky
mov ss,ax
mov sp,offset peak

mov si,0
melody:
        lea bx,notes
        mov dl,ds:[bx+si];dl = next note
        cmp dl,'K'; if K then melody ends
        je exit
        ;simple switch for notes
        cmp dl,'A'
        je A
        cmp dl,'B'
        je B
        cmp dl,'C'
        je C
        cmp dl,'D'
        je D
        cmp dl,'E'
        je E
        cmp dl,'F'
        je F
        cmp dl,'G'
        je G
        cmp dl,'H'
        je H
        mov tone,TP

        readoctave:
        lea bx,octaves;reads next octave from array
        mov cl,ds:[bx+si]
        sub cl,'0'

        shr tone,cl; double the tone octave times (tone = tone *2^octave)

        lea bx,times;read next time from array
        mov cl,ds:[bx+si]
        sub cl,'0'
        mov time,cl

        call play;
        inc si;;next index
jmp melody;play next note
;notes asignment
A: mov tone,TA
jmp readoctave
B: mov tone,TB
jmp readoctave
C: mov tone,TC
jmp readoctave
D: mov tone,TD
jmp readoctave
E: mov tone,TE
jmp readoctave
F: mov tone,TF
jmp readoctave
G: mov tone,TG
jmp readoctave
H: mov tone,TH
jmp readoctave


Progr           ends

data            segment

notes db strophe1n,strophe2n,strophe3n,'K';notes k means end of melody
octaves db strophe1o,strophe2o,strophe3o,'0';octaves, tone =(tone = frequency of note *2^octave)
times db strophe1t,strophe2t,strophe3t,'0';just times to play each note
tone dw 0
time db 0

data            ends

stacky          segment
                dw    100h dup(0)
peak          Label word
stacky          ends

end start

我知道代码有点大，但我不知道问题出在哪里。

@EDIT 也许数组大小太大，但我不这么认为@EDIT2 程序在第 78 个音符上失败，无论它是什么音符。

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2 回答 2

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在 MS-DOS 中，不要试图自己管理堆栈，除非您确切地知道自己在做什么。我唯一一次看到这样做是在一个内存非常关键的 TSR 应用程序中。

另见：为什么我们必须在 MS-DOS 中初始化 DS 和 ES 寄存器？

此外，256 字节非常小；请注意，当您的程序运行时，所有硬件中断处理程序都可能占用您的堆栈份额。

我建议你删除：

mov  ax,stacky
mov  ss,ax
mov  sp,offset peak

和

stacky  segment
        dw     100h dup(0)
peak    Label  word
stacky  ends

希望有帮助。

于 2015-01-10T18:18:02.620 回答
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原来，

interval:
        mov ah,86h;
        int 15h
        ret

工作错误，因为 int 15h 86h 函数以某种方式失败，

所以我写了自己的程序

interval:
        mov ah, 0
        int 1Ah ; actual time
        mov bx,dx
        delay:
                mov ah, 0
                int 1Ah
                sub dx,bx
                cmp di,dx
        ja delay
        ret

还注意到以其他方式定义他们的时间（现在在 di 寄存器中）例如：

note:;0,5 sec
        mov  di, 8
        call interval
        ret
halfnote:;0,25 sec
        mov  di, 4
        call interval
        ret
于 2015-01-10T19:41:26.617 回答