单片机 控制一个灯闪烁 延时2秒

Answer 1

#include<reg51.h>
sbit P_1=P1^0
void delay(int x)
{int i,j;
for(i=0;i<x;i++)
for(j=0;j<=600;j++);
}
main()
unsigned char i;
while(1)
P1=0;
delay(400);
P1=~P1;
}

Answer 2

什么型号单片机？　就用循环就可以实现啊，导通LED　，延时2S　，断开LED，延时2S，进入while（1）的死循环！

Answer 3

#include<reg51.h>
sbit led = P1^0;
void delay_xms(unsigned int x)
{
unsigned char i,j;
for (i = 0; i != x; ++i)
{
for (j = 0; j != 123; ++j);
}
}
void main()
{
while(1)
{
led=~led;
delay_xms(2000);
}
}

参考资料：随手写的 没运行过 你试试

Answer 4

#include<reg51.h>
sbit led = P1^0;
void delay_xms(unsigned int x)
{
unsigned char i,j;
for (i = 0; i != x; ++i)
{
for (j = 0; j != 123; ++j);
}
}
void main()
{
while(1)
{
delay_xms(2000);
led=~led;

}
}
/*参考楼上的。*/

Answer 5

自己探索很慢，建议买个学习板，上面带程序