51单片机学习之流水灯蜂鸣器

流水灯

在代码中的延时均使用软件延时，即函数嵌套循环体进行延时；

位输出实现

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#include<reg52.h>
#define uint unsigned int

#define on 1
#define off 0

sbit led_0 = P1 ^ 0;
sbit led_1 = P1 ^ 1;
sbit led_2 = P1 ^ 2;
sbit led_3 = P1 ^ 3;
sbit led_4 = P1 ^ 4;
sbit led_5 = P1 ^ 5;
sbit led_6 = P1 ^ 6;
sbit led_7 = P1 ^ 7;

//延时函数
void Delay(uint ms){
	uint i ,j;
	for(i = ms; i > 0; i--){
		for(j = 112; j>0; j--);
	}
}

void main(){
	while(1){	
		led_0 = on;
		Delay(200);
		led_0 = off;
        
        led_1 = on;
        Delay(200);
		led_1 = off;
        
        led_2 = on;
        Delay(200);
		led_2 = off;
        
        led_3 = on;
        Delay(200);
		led_3 = off;
        
        led_4 = on;
        Delay(200);
		led_4 = off;
        
        led_5 = on;
        Delay(200);
		led_5 = off;
        
        led_6 = on;
        Delay(200);
		led_6 = off;
        
        led_7 = on;
        Delay(200);
		led_7 = off;
	}
}

循环位移实现

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P1 = 0xfe	//1111 1110
while(1){
    Delay(1000);
    P1 = _crol_(P1,1);	//将P1左移一位
}

_crol_()函数的使用，必须要加上#include "intrins.h" ，引入头文件，该函数的原型为：

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unsigned char _crol_(unsigned char val,unsigned char n);

函数功能：以位形式将 val 左移 n 位，该函数与8051“RLA”指令相关。如果二进制数为 01010101 那么 _crol_(1) 左移1位后将高位补低位，结果10101010。

移位操作实现

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int i = 0;
while(1){
    P1 = 0xfe;
    for (i = 0; i < 8; i++){
        Delay(1000);
        P1 <<= 1;
        P1 = P1 | 0X01;
    }
}

通过每次位移一位，再与0x01进行或操作，这样每次都有一位是低电位，从而实现流水灯。

蜂鸣器