【51单片机4位数循环小数位移数值位移】2023-6-9

缘由https://ask.csdn.net/questions/7960415/54230766

大概就是在矩阵键盘输入4个数字到数码管上，然后数字再按照一定的规律进行动态变化。

可有偿代写，难度不大，stc89c52的板子

#include "reg52.h"
unsigned char code ShuMaGuan[]={0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D,0x07,0x7F,0x6F,0x00,0X80,118,121,56,63};//0~9隐.HELO
unsigned char code JZBH[]={231,215,183,119,235,219,187,123,237,221,189,125,238,222,190,126};//矩阵按要求编号15确定16清零
unsigned char duli16()
{//缘由
	unsigned char jz=255;
	static unsigned char Ys=0;
	P1=239;
	if(P1!=239&&++Ys==0){jz=P1;while(P1!=239);}
	else P1=223;
	if(P1!=223&&++Ys==0){jz=P1;while(P1!=223);}
	else P1=191;
	if(P1!=191&&++Ys==0){jz=P1;while(P1!=191);}
	else P1=127;
	if(P1!=127&&++Ys==0){jz=P1;while(P1!=127);}
	P1 = 255;
	return jz;
}
unsigned char WX4FCZ(unsigned char* a, unsigned char n, unsigned char x)
{
	unsigned char z = n / 2, j = 0; --n;
	xf:if (n - j)
	{
		if (a[j] == x)return j;
		else if (z - j < j&&a[z - j] == x)return z - j;
		else if (z + j < n&&a[z + j] == x)return z + j;
		else if (n - j > j&&a[n - j] == x)return n - j;
		else ++j;
		goto xf;
	}
	   if (a[j] == x)return j;
}
void main()
{//缘由https://ask.csdn.net/questions/7960415/54230766
	unsigned char my=0,ys=0,wei=0,x=0,jz=0,sz=0,qy=0,xs=0;
	unsigned int sz4=0;bit qd=0;
	while(1)
	{
		jz=duli16();
		if(jz!=255)
		{//获取键盘值合成
			sz=jz;
			sz=WX4FCZ(JZBH,16,sz);
			if(sz>=0&&sz<=9)sz4=sz4*10+sz;
			if(sz==14)qd=1;//确定
			if(sz==15)qd=sz4=0;//清零
		}
		P2=0;
		if(wei==0)++wei;
		P0=~wei;wei*=2;
		P2=ShuMaGuan[(x==7?sz4%10:x==6?sz4/10%10:x==5?sz4/100%10:x==4?sz4/1000:10)]
		+(xs==1&&x==6||xs==2&&x==5||xs==3&&x==4?128:0);
		if(++x>7)x=0;
		if(qd&&++my==0)if(++ys==10)
		{if(++xs>3){qy=sz4/1000;sz4%=1000;sz4=sz4*10+qy;xs=0;}ys=0;}
	}
}