怎样用stc89c52单片机控制蜂鸣器发出音乐的c语言程序？

#include\r\n#include\r\n//本例采用89C52,晶振为11.0592MHZ\r\n//关于如何编制音乐代码,其实十分简单,各位可以看以下代码.\r\n//频率常数即音乐术语中的音调,而节拍常数即音乐术语中的多少拍\r\n//所以拿出谱子,试探编吧!\r\nsbitBeep=P1^5\r\n\r\nunsignedcharn=0//n为节拍常数变量\r\nunsignedcharcodemusic_tab[]={\r\n0x18,0x30,0x1C,0x10,//格式为:频率常数,节拍常数,频率常数,节拍常数,\r\n0x20,0x40,0x1C,0x10,\r\n0x18,0x10,0x20,0x10,\r\n0x1C,0x10,0x18,0x40,\r\n0x1C,0x20,0x20,0x20,\r\n0x1C,0x20,0x18,0x20,\r\n0x20,0x80,0xFF,0x20,\r\n0x30,0x1C,0x10,0x18,\r\n0x20,0x15,0x20,0x1C,\r\n0x20,0x20,0x20,0x26,\r\n0x40,0x20,0x20,0x2B,\r\n0x20,0x26,0x20,0x20,\r\n0x20,0x30,0x80,0xFF,\r\n0x20,0x20,0x1C,0x10,\r\n0x18,0x10,0x20,0x20,\r\n0x26,0x20,0x2B,0x20,\r\n0x30,0x20,0x2B,0x40,\r\n0x20,0x20,0x1C,0x10,\r\n0x18,0x10,0x20,0x20,\r\n0x26,0x20,0x2B,0x20,\r\n0x30,0x20,0x2B,0x40,\r\n0x20,0x30,0x1C,0x10,\r\n0x18,0x20,0x15,0x20,\r\n0x1C,0x20,0x20,0x20,\r\n0x26,0x40,0x20,0x20,\r\n0x2B,0x20,0x26,0x20,\r\n0x20,0x20,0x30,0x80,\r\n0x20,0x30,0x1C,0x10,\r\n0x20,0x10,0x1C,0x10,\r\n0x20,0x20,0x26,0x20,\r\n0x2B,0x20,0x30,0x20,\r\n0x2B,0x40,0x20,0x15,\r\n0x1F,0x05,0x20,0x10,\r\n0x1C,0x10,0x20,0x20,\r\n0x26,0x20,0x2B,0x20,\r\n0x30,0x20,0x2B,0x40,\r\n0x20,0x30,0x1C,0x10,\r\n0x18,0x20,0x15,0x20,\r\n0x1C,0x20,0x20,0x20,\r\n0x26,0x40,0x20,0x20,\r\n0x2B,0x20,0x26,0x20,\r\n0x20,0x20,0x30,0x30,\r\n0x20,0x30,0x1C,0x10,\r\n0x18,0x40,0x1C,0x20,\r\n0x20,0x20,0x26,0x40,\r\n0x13,0x60,0x18,0x20,\r\n0x15,0x40,0x13,0x40,\r\n0x18,0x80,0x00\r\n}\r\n\r\nvoidint0()interrupt1//采用中断0控制节拍\r\n{TH0=0xd8\r\nTL0=0xef\r\nn--\r\n}\r\n\r\nvoiddelay(unsignedcharm)//控制频率延时\r\n{\r\nunsignedi=3*m\r\nwhile(--i)\r\n}\r\n\r\nvoiddelayms(unsignedchara)//豪秒延时子程序\r\n{\r\nwhile(--a)//采用while(--a)不要采用while(a--)各位可编译一下看看汇编结果就知道了!\r\n}\r\n\r\nvoidmain()\r\n{unsignedcharp,m//m为频率常数变量\r\nunsignedchari=0\r\nTMOD&=0x0f\r\nTMOD|=0x01\r\nTH0=0xd8TL0=0xef\r\nIE=0x82\r\nplay:\r\nwhile(1)\r\n{\r\na:p=music_tab[i]\r\nif(p==0x00){i=0,delayms(1000)gotoplay}//如果碰到结束符,延时1秒,回到开始再来一遍\r\nelseif(p==0xff){i=i+1delayms(100),TR0=0gotoa}//若碰到休止符,延时100ms,继续取下一音符\r\nelse{m=music_tab[i++],n=music_tab[i++]}//取频率常数和节拍常数\r\nTR0=1//开定时器1\r\nwhile(n!=0)Beep=~Beep,delay(m)//等待节拍完成,通过P1口输出音频(可多声道哦!)\r\nTR0=0//关定时器1\r\n}\r\n}\r\n\r\n你参考这个吧

显示子程序和T0中断程序有错。程序修改如下

#include<reg51.h>

#define uint unsigned int

#define uchar unsigned char

sbit LSA=P2^2

sbit LSB=P2^3

sbit LSC=P2^4

sbit led1=P2^0

uchar code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,

0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71}

void delayms(uint)

void display()//显示程序不用带参数

uchar shi,ge,num,num1,num2

void main()

{

TMOD=0x11

TH0=(65535-45872)/256

TL0=(65535-45872)%256

TH1=(65535-45872)/256

TH1=(65535-45872)%256

EA=1

ET0=1

ET1=1

TR0=1

TR1=1

while(1)

{

display()

}

}

void display()//显示程序不用带参数

{

shi=num/10

ge=num%10

LSA=0LSB=0LSC=0

P0=table[shi]

delayms(5)

LSA=1LSB=0LSC=0

P0=table[ge]

delayms(5)

}

void delayms(uint zms)

{

uint i,j

for(i=zmsi>0i--)

for(j=110j>0j--)

}

void T0_time()interrupt 1

{

TH0=(65535-45872)/256

TL0=(65535-45872)%256

num1++//不是num

if(num1==4)

{

num1=0led1=~led1

}

}

void T1_time()interrupt 3

{

TH1=(65535-45872)/256

TL1=(65535-45872)%256

num2++

if(num2==20)

{

num2=0num++

if(num==60) num=0

}

}

仿真结果

你是要读出stc89c52单片机里面的程序吗？

这是不可能的，STC单片机只能下载程序，即烧录程序，是不能读回程序的。

有的单片机，破解后是可以读出里面的程序的。但读出的是机器码，即二进制代码程序，需要反汇编成汇编程序。如果你对汇编程序不太懂，就是反汇编出来也没有用啊。

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