單片機(jī)eeprom如何存大于256的數(shù)

ID:650754 · 發(fā)表于 2020-2-19 15:30

想做一個(gè)帶eeprom的0-999的計(jì)數(shù)器不知道大于256的數(shù)怎么處理只會(huì)做0-99的計(jì)數(shù)器求大佬們幫忙

ID:213173 · 發(fā)表于 2020-2-19 17:17

unsigned int a,b;
unsigned char c,d;
a=999;//最大65535
//分解為2個(gè)字節(jié)保存在eeprom
c=a;//溢出部分拋棄
d=a>>8;
//讀取eeprom后合并為1個(gè)16位數(shù)據(jù)
b=d<<8|c;

ID:94031 · 發(fā)表于 2020-2-19 17:21

把大于256的數(shù)分解成高8位與低8位2個(gè)8位來存，如果用C語言直接用int類型存，編譯器給你搞定。

ID:650754 · 發(fā)表于 2020-2-19 19:26

wulin 發(fā)表于 2020-2-19 17:17
unsigned int a,b;
unsigned char c,d;
a=999;//最大65535

#include <REGX52.H>
#define uchar unsigned char
#define uint unsigned int
bit write=0;
sbit sda=P2^0;
sbit scl=P2^1;
sbit dula=P2^6;
sbit wela=P2^7;
uint sec;
uint tect;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
void delay()
{ ; ; }
void delay1(uint mun)
{
uint x,y;
   for(x=mun;x>0;x--)
for(y=110;y>0;y--);

}
void init()
{
sda=1;
delay();
scl=1;
delay();
}
void start()
{
sda=1;
delay();
scl=1;
delay();
sda=0;
delay();
}
void stop()
{
sda=0;
delay();
scl=1;
delay();
sda=1;
delay();
}
void respons()
{
uchar i;
scl=1;
delay();
while((sda==1)&&i<255)
i++;
scl=0;
delay();
}
void write_byte(uint date)
{
uchar i;
uint k;
k=date;
   for(i=0;i<8;i++)
   {
k=k<<1;
   scl=0;
delay();

sda=CY;
delay();
scl=1;
delay();
   }
   scl=0;
   delay();
   sda=1;
   delay();
}
uchar read_byte()
{
uchar i,k;
scl=0;
delay();
sda=1;
delay();
for(i=0;i<8;i++)
{
scl=1;
delay();
k=(k<<1)|sda;
delay();
scl=0;
delay();
}
//scl=1;
//delay();
//sda=1;
//delay();
return k;
}
void write_add(uchar address,uchar date)
{
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
write_byte(date);
respons();
stop();
}
uchar read_add(uchar address)
{
uchar date;
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
//delay1(5);
start();
write_byte(0xa1);
respons();
date=read_byte();
//respons();
stop();
return date;
}
void display(uint bai,uint shi,uint ge)
{
dula=0;
P0=table[bai];
dula=1;
dula=0;
wela=0;
P0=0x7e;
wela=1;
wela=0;
delay1(5);

dula=0;
P0=table[shi];
dula=1;
dula=0;
wela=0;
P0=0x7d;
wela=1;
wela=0;
delay1(5);

dula=0;
P0=table[ge];
dula=1;
dula=0;
wela=0;
P0=0x7b;
wela=1;
wela=0;
delay1(5);
}
void main()
{
init();
sec=read_add(2)+256*read_add(3);
if(sec>1000)
sec=0;
TMOD=0x01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
EA=1;
ET0=1;
TR0=1;
while(1)
{
display(sec/100,sec%100/10,sec%10);
if(write==1)
{
write=0;

write_add(2,sec);
//delay1(100);
write_add(3,sec>>8);
}
}
}
void t0() interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;

tect++;
if(tect==1)
{
tect=0;
sec++;
write=1;
if(sec==1000)
sec=0;
}
}
你看看我這樣子寫哪里有問題？

ID:650754 · 發(fā)表于 2020-2-19 19:27

xuyaqi 發(fā)表于 2020-2-19 17:21
把大于256的數(shù)分解成高8位與低8位2個(gè)8位來存，如果用C語言直接用int類型存，編譯器給你搞定。

#include <REGX52.H>
#define uchar unsigned char
#define uint unsigned int
bit write=0;
sbit sda=P2^0;
sbit scl=P2^1;
sbit dula=P2^6;
sbit wela=P2^7;
uint sec;
uint tect;
uchar code table[]={
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71};
void delay()
{ ; ; }
void delay1(uint mun)
{
uint x,y;
   for(x=mun;x>0;x--)
for(y=110;y>0;y--);

}
void init()
{
sda=1;
delay();
scl=1;
delay();
}
void start()
{
sda=1;
delay();
scl=1;
delay();
sda=0;
delay();
}
void stop()
{
sda=0;
delay();
scl=1;
delay();
sda=1;
delay();
}
void respons()
{
uchar i;
scl=1;
delay();
while((sda==1)&&i<255)
i++;
scl=0;
delay();
}
void write_byte(uint date)
{
uchar i;
uint k;
k=date;
   for(i=0;i<8;i++)
   {
k=k<<1;
   scl=0;
delay();

sda=CY;
delay();
scl=1;
delay();
   }
   scl=0;
   delay();
   sda=1;
   delay();
}
uchar read_byte()
{
uchar i,k;
scl=0;
delay();
sda=1;
delay();
for(i=0;i<8;i++)
{
scl=1;
delay();
k=(k<<1)|sda;
delay();
scl=0;
delay();
}
//scl=1;
//delay();
//sda=1;
//delay();
return k;
}
void write_add(uchar address,uchar date)
{
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
write_byte(date);
respons();
stop();
}
uchar read_add(uchar address)
{
uchar date;
start();
write_byte(0xa0);
respons();
write_byte(address);
respons();
//delay1(5);
start();
write_byte(0xa1);
respons();
date=read_byte();
//respons();
stop();
return date;
}
void display(uint bai,uint shi,uint ge)
{
dula=0;
P0=table[bai];
dula=1;
dula=0;
wela=0;
P0=0x7e;
wela=1;
wela=0;
delay1(5);

dula=0;
P0=table[shi];
dula=1;
dula=0;
wela=0;
P0=0x7d;
wela=1;
wela=0;
delay1(5);

dula=0;
P0=table[ge];
dula=1;
dula=0;
wela=0;
P0=0x7b;
wela=1;
wela=0;
delay1(5);
}
void main()
{
init();
sec=read_add(2)+256*read_add(3);
if(sec>1000)
sec=0;
TMOD=0x01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
EA=1;
ET0=1;
TR0=1;
while(1)
{
display(sec/100,sec%100/10,sec%10);
if(write==1)
{
write=0;

write_add(2,sec);
//delay1(100);
write_add(3,sec>>8);
}
}
}
void t0() interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;

tect++;
if(tect==1)
{
tect=0;
sec++;
write=1;
if(sec==1000)
sec=0;
}
}
我這樣子寫哪里出現(xiàn)問題了

ID:691440 · 發(fā)表于 2020-2-19 19:45

強(qiáng)制類型轉(zhuǎn)換符可以幫你做到。取高8位，低8位。

ID:57657 · 發(fā)表于 2020-7-25 20:19

能用union的地方就不要用移位，影響效率

#include "reg51.h"
void main(){
union {
unsigned char c[4];
unsigned long l; //最大4294967295
} b;
b.l = 0xABCD1234;
P0 = b.c[0];
P1 = b.c[1];
P2 = b.c[2];
P3 = b.c[3];
while (1);
}

復(fù)制代碼

ID:808700 · 發(fā)表于 2020-7-28 11:41

拆成2個(gè)8位，或者3個(gè)，4個(gè)，看你數(shù)據(jù)大小要求

ID:275449 · 發(fā)表于 2020-7-28 11:50

拆分啊就像時(shí)分秒最大不就59 但是告訴你 20:58:50 你能算成秒吧告訴你 3650秒你能換算成時(shí)吧

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單片機(jī)eeprom如何存大于256的數(shù)