標(biāo)題:
stc8f2k08s2單片機(jī)雙串口同時(shí)通訊,怎么實(shí)現(xiàn)
[打印本頁]
作者:
priam
時(shí)間:
2018-6-25 21:42
標(biāo)題:
stc8f2k08s2單片機(jī)雙串口同時(shí)通訊,怎么實(shí)現(xiàn)
stc8f2k08s2 說明上寫著支持雙串口。我在stc-isp上看到的范例單獨(dú)串口1實(shí)現(xiàn),和串口2實(shí)現(xiàn)。 這里需求需要 數(shù)據(jù) 通過串口1接收 ,在通過串口2發(fā)出去。
我想把下面的的共能和在一起,同時(shí)支持串口1的收發(fā),串口2的收發(fā)。 上面的兩段代碼 就 AUXR 的值不同。合在一起AUXR沖突。
向大家請教。 雙串口功能同時(shí)實(shí)現(xiàn),代碼怎么寫
stc-isp上串口1代碼:
----------------------------
#include "reg51.h"
#include "intrins.h"
#define FOSC 11059200UL
#define BRT (65536 - FOSC / 9600 / 4)
sfr AUXR = 0x8e;
sfr T2H = 0xd6;
sfr T2L = 0xd7;
sfr S2CON = 0x9a;
sfr S2BUF = 0x9b;
sfr IE2 = 0xaf;
bit busy;
char wptr;
char rptr;
char buffer[16];
void Uart2Isr() interrupt 8 using 1
{
if (S2CON & 0x02)
{
S2CON &= ~0x02;
busy = 0;
}
if (S2CON & 0x01)
{
S2CON &= ~0x01;
buffer[wptr++] = S2BUF;
wptr &= 0x0f;
}
}
void Uart2Init()
{
S2CON = 0x50;
T2L = BRT;
T2H = BRT >> 8;
AUXR = 0x14;
wptr = 0x00;
rptr = 0x00;
busy = 0;
}
void Uart2Send(char dat)
{
while (busy);
busy = 1;
S2BUF = dat;
}
void Uart2SendStr(char *p)
{
while (*p)
{
Uart2Send(*p++);
}
}
void main()
{
Uart2Init();
IE2 = 0x01;
EA = 1;
Uart2SendStr("Uart Test !\r\n");
while (1)
{
if (rptr != wptr)
{
Uart2Send(buffer[rptr++]);
rptr &= 0x0f;
}
}
}
--------------------結(jié)束-以上代碼可以運(yùn)行--------------------------------
stc-isp上串口2代碼:
#include "reg51.h"
#include "intrins.h"
#define FOSC 11059200UL
#define BRT (65536 - FOSC / 115200 / 4)
sfr AUXR = 0x8e;
bit busy;
char wptr;
char rptr;
char buffer[16];
void UartIsr() interrupt 4 using 1
{
if (TI)
{
TI = 0;
busy = 0;
}
if (RI)
{
RI = 0;
buffer[wptr++] = SBUF;
wptr &= 0x0f;
}
}
void UartInit()
{
SCON = 0x50;
TMOD = 0x00;
TL1 = BRT;
TH1 = BRT >> 8;
TR1 = 1;
AUXR = 0x40;
wptr = 0x00;
rptr = 0x00;
busy = 0;
}
void UartSend(char dat)
{
while (busy);
busy = 1;
SBUF = dat;
}
void UartSendStr(char *p)
{
while (*p)
{
UartSend(*p++);
}
}
void main()
{
UartInit();
ES = 1;
EA = 1;
UartSendStr("Uart Test !\r\n");
while (1)
{
if (rptr != wptr)
{
UartSend(buffer[rptr++]);
rptr &= 0x0f;
}
}
}
復(fù)制代碼
----------------------結(jié)束-以上代碼可以運(yùn)行---------------------
作者:
priam
時(shí)間:
2018-6-26 13:09
沒有問題了,問題解決了
作者:
1103278135
時(shí)間:
2018-8-13 01:53
請問怎么解決的呀?我正好也遇到 謝謝
作者:
alanj
時(shí)間:
2018-8-15 14:51
怎么解決的?程序還能發(fā)上來分享一下?
作者:
alanj
時(shí)間:
2018-8-15 14:54
請問怎么解決的?遇到這個(gè)問題!
作者:
angmall
時(shí)間:
2018-8-15 18:52
給你改了一下試試。
#include "reg51.h"
#include "intrins.h"
#define FOSC 11059200UL
#define BRT (65536 - FOSC / 115200 / 4)
#define BRT2 (65536 - FOSC / 9600 / 4)
sfr AUXR = 0x8e;
sfr T2H = 0xd6;
sfr T2L = 0xd7;
sfr S2CON = 0x9a;
sfr S2BUF = 0x9b;
sfr IE2 = 0xaf;
bit busy;
char wptr;
char rptr;
char buffer[16];
bit busy2;
char wptr2;
char rptr2;
char buffer2[16];
void UartIsr() interrupt 4 using 1
{
if (TI)
{
TI = 0;
busy = 0;
}
if (RI)
{
RI = 0;
buffer2[wptr2++] = SBUF;
wptr2 &= 0x0f;
}
}
void Uart2Isr() interrupt 8 using 1
{
if (S2CON & 0x02)
{
S2CON &= ~0x02;
busy2 = 0;
}
if (S2CON & 0x01)
{
S2CON &= ~0x01;
buffer[wptr++] = S2BUF;
wptr &= 0x0f;
}
}
void UartInit()
{
SCON = 0x50;
TMOD = 0x00;
TL1 = BRT;
TH1 = BRT >> 8;
TR1 = 1;
AUXR |= 0x40;
wptr = 0x00;
rptr = 0x00;
busy = 0;
}
void Uart2Init()
{
S2CON = 0x50;
T2L = BRT2;
T2H = BRT2 >> 8;
AUXR |= 0x14;
wptr2 = 0x00;
rptr2 = 0x00;
busy2 = 0;
}
void UartSend(char dat)
{
while (busy);
busy = 1;
SBUF = dat;
}
void Uart2Send(char dat)
{
while (busy);
busy2 = 1;
S2BUF = dat;
}
void UartSendStr(char *p)
{
while (*p)
{
UartSend(*p++);
}
}
void Uart2SendStr(char *p)
{
while (*p)
{
Uart2Send(*p++);
}
}
void main()
{
UartInit();
Uart2Init();
IE2 = 0x01;
ES = 1;
EA = 1;
UartSendStr("Uart Test !\r\n");
Uart2SendStr("Uart2 Test !\r\n");
while (1)
{
if (rptr != wptr)
{
UartSend(buffer[rptr++]);
rptr &= 0x0f;
}
if (rptr2 != wptr2)
{
Uart2Send(buffer2[rptr2++]);
rptr2 &= 0x0f;
}
}
}
//--------------------結(jié)束-以上代碼可以運(yùn)行--------------------------------
/*
void main()
{
while (1)
{
}
}
*/
復(fù)制代碼
作者:
zfeng
時(shí)間:
2018-8-16 09:26
這么多人需要,分享一下讓更多的人少走彎路!
作者:
alanj
時(shí)間:
2018-8-21 16:26
有沒有遇到采用中斷方式每隔幾秒種發(fā)送一組數(shù)據(jù)出去之后,調(diào)試助手收到多個(gè)同樣的數(shù)據(jù),而在串口發(fā)送后加1秒的延遲程序,收到的就只有一組數(shù)據(jù).
#include "stc15.h"
#include "USART.h"
#include "intrins.H"
unsigned char TxBusy = 0; //發(fā)送忙檢測
unsigned char rxBuff[rxLength]={0}; //接收緩沖區(qū),長度50
unsigned char txBuff[txLength]={0}; //發(fā)送緩沖區(qū),長度50
unsigned char rxcnt = 0; //接收計(jì)數(shù)
/* void Uart_Init(void) //115200bps@11.0592MHz
{
SCON = 0x50; //8位數(shù)據(jù),可變波特率
AUXR |= 0x01; //串口1選擇定時(shí)器2為波特率發(fā)生器
AUXR |= 0x04; //定時(shí)器2時(shí)鐘為Fosc,即1T
T2L = 0xE8; //設(shè)定定時(shí)初值
T2H = 0xFF; //設(shè)定定時(shí)初值
AUXR |= 0x10; //啟動(dòng)定時(shí)器2
ES = 1; //使能串口中斷
EA = 1; //開總中斷
}*/
void Uart_Init(void) //9600bps@11.0592MHz
{
SCON = 0x50; //8位數(shù)據(jù),可變波特率
AUXR |= 0x01; //串口1選擇定時(shí)器2為波特率發(fā)生器
AUXR |= 0x04; //定時(shí)器2時(shí)鐘為Fosc,即1T
T2L = 0xE0; //設(shè)定定時(shí)初值
T2H = 0xFE; //設(shè)定定時(shí)初值
AUXR |= 0x10; //啟動(dòng)定時(shí)器2
ES = 1; //使能串口中斷
EA = 1; //開總中斷
}
/*
void Uart1_SendByte(unsigned char dat)
{
//unsigned int OutTime = 10000;
//TI = 0;
//while((TxBusy)&&(OutTime--));
while(TxBusy);
SBUF = dat;
TxBusy = 1;
}*/
void Uart1_SendByte(unsigned char dat)
{
while(TxBusy);//等待數(shù)據(jù)發(fā)送完成,進(jìn)入中斷,將TI、TxBusy清0
TxBusy = 1;
SBUF = dat;
}
void Uart1_SendByte_N(unsigned char *dat,unsigned char N)
{
while(N--)
{
Uart1_SendByte(*dat) ;
dat++;
}
}
void Uart1_IT_Handle(void) interrupt 4 using 1
{
unsigned char temp;
if(RI)
{
RI = 0; //清RI
if(rxcnt<rxLength)
{
rxBuff[rxcnt++]= SBUF;
}
else
{
temp =SBUF; //讀BUFF
}
}
if(TI)
{
TI = 0 ; //清TI
TxBusy = 0 ; //清BUSY
}
}
/*void Uart_SendStr(const char* str)
{
while(*str)
{
Uart1_SendByte(*str);
str++;
}
} */
作者:
米明
時(shí)間:
2019-9-3 21:20
請問是怎么實(shí)現(xiàn)的呢?
作者:
111111486
時(shí)間:
2019-11-27 14:17
為什么我把例程刷進(jìn)去發(fā)送的是11但是返回卻是19
51hei瀏覽器截圖20191127141449.png
(21.91 KB, 下載次數(shù): 47)
下載附件
2019-11-27 14:16 上傳
作者:
聰明的喜羊羊
時(shí)間:
2019-11-27 14:23
請問怎么解決的,謝謝
作者:
聰明的喜羊羊
時(shí)間:
2019-11-27 14:24
這么多人需要,分享一下讓更多的人少走彎路。
歡迎光臨 (http://www.torrancerestoration.com/bbs/)
Powered by Discuz! X3.1