有人可以帮我吗? 我编写了下面的代码来运行数字时钟,但是当我在 proteus 中进行模拟时,LED 只闪烁,显示屏只显示 00:00,而且永远不会计数。我使用定时器 0 和定时器 1 来生成 1 秒的延迟
有什么建议吗?
函数和变量
int display(int num);
int clock_PIC();
/*VARIAVEIS GLOBAIS*/
char controlador = 0x01;
int unidade, dezenas, centenas, milhares;
char segundos = 0x00;
char minutos = 0x00;
char horas = 0x00;
char clck_cont = 0x00;
char flags_min = 0x00;
char flags_hor = 0x00;
中断(定时器0和定时器1)
void interrupt interrupcao ()
{
// int a;
// display(a);
if(INTCONbits.TMR0IF)
{
if(!digito_milhares && controlador == 1)
{
controlador = 0x02;
digito_unidade = 0x00;
digito_dezenas = 0x00;
digito_centenas = 0x00;
PORTC = 0x00;
milhares = (horas%100)/10;
digito_milhares = 0x01;
PORTC = display(milhares);
}//end if digito milhares
else if(!digito_centenas && controlador == 2)
{
controlador = 0x03;
digito_unidade = 0x00;
digito_dezenas = 0x00;
digito_milhares = 0x00;
PORTC = 0x00;
centenas = horas%10;
digito_centenas = 0x01;
PORTC = display(centenas);
}//end if digito centenas
else if(!digito_dezenas && controlador == 3)
{
controlador = 0x04;
digito_unidade = 0x00;
digito_centenas = 0x00;
digito_milhares = 0x00;
PORTC = 0x00;
dezenas = (minutos%100)/10;
digito_dezenas = 0x01;
PORTC = display(dezenas);
}//end if digito dezenas
else if(!digito_unidade && controlador == 4)
{
controlador = 0x01;
digito_dezenas = 0x00;
digito_centenas = 0x00;
digito_milhares = 0x00;
PORTC = 0x00;
unidade = minutos%10;
digito_unidade = 0x01;
PORTC = display(unidade);
}//end if digito unidade
// clear the TMR0 interrupt flag
INTCONbits.TMR0IF = 0x00;
}//end if TMR0IF
if(PIR1bits.TMR1IF)
{
LATBbits.LATB7 = ~ LATBbits.LATB7;
clck_cont ++;
if(clck_cont == 0x02)
{
clck_cont = 0x00;
segundos++;
}//end clck_cont
// clear the TMR1 interrupt flag
PIR1bits.TMR1IF = 0x00;
//TMR1H 11;
TMR1H = 0x0B;
//TMR1L 220;
TMR1L = 0xDC;
}//end TMR1IF
}//end interrupcao
主要功能
void main(void)
{
PORTB = 0x4F;
PORTC = 0x3F;
/**
TRISx registers
*/
TRISB = 0x70;
TRISC = 0x80;
/**
ANSELx registers
*/
ANSELC = 0x00;
ANSELB = 0x00;
// SCS FOSC; IRCF 4MHz_HFINTOSC/4; IDLEN disabled;
OSCCON = 0x50;
// PRISD enabled; SOSCGO disabled; MFIOSEL disabled;
OSCCON2 = 0x04;
// INTSRC disabled; PLLEN disabled; TUN 0;
OSCTUNE = 0x00;
// TMR0H 0;
TMR0H = 0x00;
// TMR0L 6;
TMR0L = 0x06;
// T0PS 1:16; T08BIT 8-bit; T0SE Increment_hi_lo; T0CS FOSC/4; TMR0ON enabled; PSA assigned;
T0CON = 0xD3;
//T1GSS T1G_pin; TMR1GE disabled; T1GTM disabled; T1GPOL low; T1GGO done; T1GSPM disabled;
T1GCON = 0x00;
//TMR1H 11;
TMR1H = 0x0B;
//TMR1L 220;
TMR1L = 0xDC;
// Clearing IF flag before enabling the interrupt.
// PIR1bits.TMR1IF = 0;
// T1CKPS 1:8; T1OSCEN disabled; T1SYNC do_not_synchronize; TMR1CS FOSC/4; TMR1ON enabled; T1RD16 enabled;
T1CON = 0x37;
INTCONbits.GIE = 1; //Habilita a interrupcao global
PIE1bits.TMR1IE = 1; //Habilita a interrupcao do timer 1
INTCONbits.TMR0IE = 1; //Habilita a interrupcao do timer 0
INTCON2bits.nRBPU = 0; //Habilita o pull up interno do port B
while (1)
{
clock_PIC();
}//end while
}//end void
时钟功能
int clock_PIC()
{
if (segundos > 59)
{
segundos = 0x00;
minutos++;
if(minutos > 59)
{
minutos = 0x00;
horas++;
if (horas > 23)
{
horas = 0x00;
}//end horas
}//end minutos
}//end segundos
if(!butt_minutos) flags_min = 0x01;
if(!butt_horas) flags_hor = 0x01;
if(butt_minutos && flags_min)
{
flags_min = 0x00;
minutos++;
if(minutos > 59) minutos = 0x00;
}
if(butt_horas && flags_hor)
{
flags_hor = 0x00;
horas++;
if(horas > 23) horas = 0x00;
}
}//end clock_PIC
显示功能
int display(int num)
{
int cathode; //armazena código BCD
// -- Vetor para o código BCD --
int SEGMENTO[10] = {0x3F, //BCD zero '0'
0x06, //BCD um '1'
0x5B, //BCD dois '2'
0x4F, //BCD três '3'
0x66, //BCD quatro '4'
0x6D, //BCD cinco '5'
0x7D, //BCD seis '6'
0x07, //BCD sete '7'
0x7F, //BCD oito '8'
0x67}; //BCD nove '9'
cathode = SEGMENTO[num]; //para retornar o cathode
return(cathode); //retorna o número BCD
} //end display
最佳答案
我真的不知道你正在使用的具体PIC,但作为嵌入式系统的一般规则,中断应该由简短的代码组成。中断机制将调用该代码,并且预计会在短时间内完成,因为另一个中断可能会出现并被错过(即使是同一个 ISR)。因此,常见的做法是在 ISR 回调内部进行一些处理(例如,将中断标记为已处理)并更改标记已生成中断的标志,并在 main()
中完成大部分工作。或其他功能。这并不总是可能的,因为这有点像池化而不是使用回调,但这对于低端微 Controller 来说很常见。
除了对于 ISR 来说较大的代码之外,您还应该避免在中断回调中调用其他函数,因为它会增加开销。此外,在这种特殊情况下,您正在使用模运算符 %
,这意味着除法:如果 PIC 没有硬件除法器,则操作速度非常慢。
您要做的就是移出 ISR void interrupt interrupcao ()
内的代码到另一个函数,只留下必须更改其中寄存器的代码,并标记中断已发生。然后检查主循环内的该标志并进行处理。
当中断回调很长时,另一个曾经把一切搞乱的事情是分配给函数的空间,即物理内存。使其变大会使代码溢出到另一个中断空间;但我怀疑这种情况发生在你身上,因为我只看到它在汇编中编码(编译器会将代码放在正确的位置,但我不知道你是如何编译的;))。
关于c++ - 带 pic18f445k22 的 7 段显示数字时钟,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56814042/