type of data

Before writing the running light program, let's understand the data types.

When using corresponding data, you should pay attention to overflow problems. If there is overflow, the compiler may compile it through, but the execution result will be wrong.

Flowing lights

Before operating the LED, you must first understand what level the LED is lit at and what pins the LED is connected to. Usually you need to look up the schematic diagram. Here my 8 LEDs are connected to port P2 and light up at a low level (because the ability to sink current is stronger than the ability to source current).

1. Functional cycle water light:

#include "reg52.h"   

#include //Function header file

#define led P2 //Define the P2 port of the entire LED

/* Delay function */

void delay(unsigned int del) //10ms * del

{

unsigned int i,j;

for(i=0; i for(j=0; j<1827; j++);

}

/* while(i--);//delay 10 * i us; because a single instruction is 1us; */

void main()

{

unsigned char i; //Use char to save RAM

led=0xfe; //Initial state turns on LED0, 1111 1110

while(1)

{

for(i=0;i<7;i++) //Shift the led one position to the left in a loop

{

led=_crol_(led,1);

delay(10); // Delay about 100ms, LED display is faster

}

// At this time, the LED is 0111 1111

for(i=0;i<7;i++) //Move the LED right one bit in a loop

{

led=_cror_(led,1);

delay(10); 

}

}

}

Logical shift (fill 0) running light

#include "reg52.h"   

#include //This is the function header file

#define led P2 //Operate the entire P2 port

/* Delay function */

void delay(unsigned int del) //10ms * del

{

unsigned int i,j;

for(i=0; i for(j=0; j<1827; j++)      

    ;

}

/* while(i--);//Delay of about 10us; because a single instruction takes 1us; */

void main()

{

unsigned  char i;

led=0xfe; //Initial state turns on LED0

while(1)

{

for(i=0;i<8;i++)// 1111 1110

{

P2=~(0x01< delay(10); // Delay about 100ms, quite fast

}

for(i=0;i<8;i++)

{

P2=~(0x80>>i); //Shift 1 right by i bits, then invert the result and assign it to port P2

delay(10); 

}

}

}

The two methods have the same effect. So far, the LED operation is complete.