//115.2k串行口通信程序，校驗(yàn)方式為簡(jiǎn)單雙向校驗(yàn),
//而且是必要的，因?yàn)槿魏尾捎弥袛?、crc等的方法
//均很難保證速度和避免錯(cuò)誤.
//總體速度5KB/秒

//單片機(jī)程序
//使用22.1184晶振
unsigned char time,b_break,b_break_3;
//35.555ms
void int_t0(void) interrupt 1 {
?TH0=0;
?if ((++b_break_3)>2) b_break=1;
?time++;
}

//

unsigned char rec(void) {
?TH0=0;b_break_3=0;b_break=0;
?while (RI==0) {
? if (b_break) return(1);? //等待pc發(fā)送100ms超時(shí)
?}
?RI=0;
?SBUF=SBUF;
?return(SBUF);
}
/*返回1表示失敗*/
bit sen(unsigned char d) {
?SBUF=d;
?TH0=0;b_break_3=0;b_break=0;
?while (RI==0) {
? if (b_break) return(1);? //等待pc送回相同字節(jié)數(shù)據(jù)，100ms超時(shí)
?}
?RI=0;
?if (SBUF!=d) return(1);
?return(0);
}

void main(void) {
?unsigned int i;
?unsigned char d;
?EA=1;
?SCON=0xd8;PCON=0x80;
?TMOD=0x21;
?TL1=TH1=0xff;TR1=1;
?TH0=0;ET0=TR0=1;
?while (1) {
? d=rec();
? if ((!b_break)&&(d==0x55)) {
?? d=rec();
?? if (!b_break) {
??? if (d==0xaa) {
???? //接收到pc送來(lái)的要求發(fā)送指令0xaa55
???? for (i=0;i<1000;i++) {
????? if (sen(i)) {
?????? //發(fā)送錯(cuò)誤
?????? break;
????? }
???? }
??? } else if (dd==0xaa) {
???? //接收到pc送來(lái)的要求接收指令0xaaaa
???? for (i=0;i<1000;i++) {
????? d=rec();
????? if (!b_break) {
?????? //正確接收處理
????? } else {
?????? //接收錯(cuò)誤處理
?????? break;
????? }
???? }
??? }
?? }
? }
?}
}

//電腦程序tc2.0
＃i nclude "stdio.h"
＃i nclude "dos.h"
＃i nclude "process.h"
＃i nclude "string.h"
＃i nclude "conio.h"
＃i nclude

unsigned char time_out;

void init_com1(void) {
?outportb(0x3fb,0x80);
?outportb(0x3f8,1);
?outportb(0x3f9,0x0);
?outportb(0x3fb,0x3);
}
void sen_com1(unsigned char d) {
?long i;
?time_out=0;
?while (!(inportb(0x3fd)&0x20));
?outportb(0x3fd,0x0);
?outportb(0x3f8,d);
?i=clock();
?while (1) {
? if (clock()>(i+8)) {time_out=1;return;}
? if ((inp(0x3fd)&0x1)==1) break;
?}
?if (inp(0x3f8)!=d) {time_out=2;return;}
}
unsigned char rec_com1(void) {
?long i;
?unsigned char d;
?time_out=0;
?i=clock();
?while (1) {
? if (clock()>(i+8)) {time_out=1;return(0);}
? if ((inp(0x3fd)&1)==1) break;
?}
?d=inp(0x3f8);
?while (!(inportb(0x3fd)&0x20));
?outportb(0x3fd,0x0);
?outportb(0x3f8,d);
?return(d);
}
void main(void) {
?int i;
?long st;
?init_com1();
?st=clock();
?sen_com1(0xaa);
?if (time_out) {printf("CPU not ready!");exit(-1);}
?sen_com1(0xaa);
?if (time_out) {printf("CPU not ready!");exit(-1);}
?for (i=0;i<1000;i++) {
? sen_com1(i);
? if (time_out) {printf("CPU not ready!");exit(-1);}
?}
?printf("Total use %3.2f S.",(clock()-st)/18.2);
?exit(1);
}