单片机~~用八个发光二极管实现8位二进制的累加

Question

具体程序如下 ：

#include<reg52.h>
#define uchar unsigned char
#define uint unsigned int
void delay(uint z);
void main()
{
uchar temp;
****temp=0xff;
while(1)
{
temp++;
P1=~temp;
delay(1000);
}*****
}
void delay(uint z)
{
uint x,y;
for(x=z;x>0;x--)
for(y=110;y>0;y--);
}

我有几个地方没看懂 那个***.....***之间的看不大明白 过程是什么样的啊？？谢谢各位帮忙~~~~！！！

Answer 1

#include "reg51.h"
void delay(unsigned int ms)
{
unsigned char i;
while(ms--) for(i = 110; i-- ; i > 0);
}
//-------------------------
void main()
{
unsigned char abc = 0;
while(1) {
P1 = abc; //在P1口外接8个LED，低电平发光.
abc++; //加，负逻辑将显示为自减.
delay(500); //延时，供观察现象.
} }
=============================================
要求由最高位P0^0作为二进制的最高位。
这个新要求，可以在焊接LED器件的时候进行调整。
更改一下程序，也不难，如下：
#include "reg51.h"
unsigned char bdata abc = 0, bbb;
sbit a0 = abc^0;
sbit a1 = abc^1;
sbit a2 = abc^2;
sbit a3 = abc^3;
sbit a4 = abc^4;
sbit a5 = abc^5;
sbit a6 = abc^6;
sbit a7 = abc^7;

sbit b0 = bbb^0;
sbit b1 = bbb^1;
sbit b2 = bbb^2;
sbit b3 = bbb^3;
sbit b4 = bbb^4;
sbit b5 = bbb^5;
sbit b6 = bbb^6;
sbit b7 = bbb^7;

void delay(unsigned int ms)
{
unsigned char i;
while(ms--) for(i = 110; i-- ; i > 0);
}

void main()
{
while(1) {
b0 = a7;
b1 = a6;
b2 = a5;
b3 = a4;
b4 = a3;
b5 = a2;
b6 = a1;
b7 = a0;
P1 = bbb;
// P1 = abc; //在P1口外接8个LED，低电平发光.
abc++; //加，负逻辑将显示为自减.
delay(500); //延时，供观察现象.
} }

Answer 2

没见过，把它去掉，我想没有什么意义