Keil UART interrupt getting hangs in printf() function

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Program hangs in printf().

C

void UART0_ISR(void) interrupt 4
{
    if (RI == 1)
    {
            RI = 0;
            TI = 1;
            Count = 444;
            datau[temp] = SBUF;             // Read receive data
            SBUF = 0;// datau[temp];           // Send back same data on uart
            Count=1;
            if(temp < 10)
            {
                temp =temp+1;
            }
            if(temp >10)
            {
                temp=0;
            }
            uartreceive=1;
            Count = 444;
        }
    else TI = 0;
}

void main() {

        AUXR1|=0x00;
        PCON&=0x7F;
        SCON  = 0x50;         // SCON: mode 1, 8-bit UART, enable rcvr
        TMOD&=0xAF;
            TMOD |= 0x20;               // TMOD: timer 1, mode 2, 8-bit reload
        TH1   = 0xFA ;                // TH1:  reload value for 1200 baud @ 16MHz
            TR1   = 1;                  // TR1:  timer 1 run
            TI    = 1;                  // TI:   set TI to send first char of UART
        ES = 1;                             // Enable serial interrupt
        EA = 1;
        Count=123;
        temp=0;
    while(1)
    {
        Delay_ms(100);
        Count=102;
        printf("sss\n %u",Count); //My program is getting hangs here

        if(uartreceive==1)
        {
            uartreceive=0;
            Count=4;
            printf("Entire array is %s  %u\n",datau,temp);
            Count=105;
            printf("\nindex  %u %c  %c\n",temp,datau[0],datau[1]);
            Count=106;
            sprintf(datau," ");
            Count=107;
            temp=0;
            uartreceive=0;
            Count=104;
        }

        }
     }

What I have tried:

I am trying to communicate from PC using UART. Its is working but stuck at the printf() function inside the while loop. I don't know what is the mistake I have done.

Posted 19-Jul-16 21:36pm

S.Soundar

Updated 6-Jun-23 20:07pm

CPallini

v2

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Comments

barneyman 20-Jul-16 5:52am

you're in the middle of an interrupt service routine - you have to get out as quickly as possible,and make no blocking calls

push the data into a pre-allocated buffer, and let another, normal, thread do something with it

S.Soundar 21-Jul-16 1:45am

I checked that sir. But interrupt is working fine. If I take out the Delay_ms(100) function its working fine. But if I am using delay program hangs on printf().

Here is my delay function code:

void Delay_us(unsigned int us_count)
{
while(us_count!=0)
{
us_count--;

}
}

void Delay_ms(unsigned int ms_count)
{
while(ms_count!=0)
{
Delay_us(550); //delay_us is called to generate 1ms delay
ms_count--;
Count=ms_count;
}

}

barneyman 21-Jul-16 2:52am

you need some form of read/write locking on that data buffer - currently it has the potential to change while printf is traversing it

[no name] 25-Jul-16 15:26pm

Does the printf use the UART to communicate to the host PC?

S.Soundar 26-Jul-16 5:12am

Yes bling.

[no name] 26-Jul-16 12:29pm

That is likely the problem. You took over the UART's ISR.
If the device has two UARTs - use one for debugging / trace via printf and the other for your application.

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Charlie Chang 2022 · Answer 1 · 2023-06-06T15:43:00

UART interrupt routiing set TI = 0 will cause printf crashed, I use sprintf and below uart_sendstring to replace printf to avoid crash. If didn't set TI = 0 in UART interrupt routing, It will cause the UART interrupt to continually occur at the frequency of the UART baud rate, causing the CPU's computing power to be wasted in handling a large number of UART interrupts.

uint8_t xdata pbf[100];
sprintf(pbf, "\033[2J\033[H");
uart_sendstring(pbf, strlen(pbf));

//---------------------------------------------------------------
void Send_Data_To_UART0 (UINT8 c)
{
TI = 0;
SBUF = c;
while(TI==0);
}

//---------------------------------------------------------------
void uart_sendstring(uint8_t* s, uint8_t cont)
{
while (cont) {
Send_Data_To_UART0(*s++);
cont--;
}
}