程序如下只需要大哥們分析一下第一個為什出了問題即可。個人認為是數字太大超過范圍不過usigned long在單片機中范圍是2的32次方。∷哉f暫時找不出問題所在。
#include<reg52.h>
#define uchar unsigned char
#define uint unsigned int
#define ulong unsigned long;
sbit wela=P2^7;
uchar num;
uint ge,shi,bai,qian,wan,sw;
sbit dula=P2^6;
ulong num2;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
/*1111 1110=0xfe 1
1111 1101=0xfd 2
1111 1011=0xfb 3
1111 0111=0xf7 4
1110 1111=0xef 5
1101 1111=0xdf 6
1011 1111=0xbf 7
1100 1111=0xcf
1100 0111=0xc7
1100 0011=0xc3
1100 0001=0xc1
1100 0000=0xc0*/
void delay(uint z)
{
uint x,y;
for(x=110;x>0;x--);
for(y=z;y>0;y--);
}
void display(uint ge,uint shi,uint bai,uint sw,uint qian,uint wan)
{
ge=num2%100000;
shi=num2%100000/10000;
bai=num2%100000%10000/1000;
qian=num2%100000%10000%1000/100;
wan=num2/60000;
sw=num2/90000;
dula=1;
P0=table[ge];
dula=0;
wela=1;
P0=0xdf;
wela=0;
delay(65000);
}
void main()
{
dula=0;
wela=0;
num=0xfe;
num2=187732;
while(1)
{
display(ge,shi,bai,qian,wan,sw);
}
}
現在只是顯第一個數字,拜托了。
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