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???
12/01/08 18:36
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#160462 - Where is the UART ISR?
Responding to: ???'s previous message
Watchdog is disabled at the main(). Surely, there is still a chance that the watchdog fires before the MCU execution reaches to the main(). But he said it goes to "EA = 1;" over the watchdog disable (PCA0MD &= ~0x40;).

Rather, the UART interrupt is enabled but no UART ISR is supplied.
void Timer1_Init (void) 
{
    ...
    ES0 = 1; // Enable UART0 interrupts 
}

void main (void)
{
    ...
    PCA0MD &= ~0x40; // WDTE = 0 (clear watchdog timer enable) 
    ...
    Timer1_Init();
    EA = 1;
    TI0=1;
}


Another point is,
- printf is used but no putchar() is implemented.

Tsuneo

List of 21 messages in thread
TopicAuthorDate
Program hangs at one point - 8051F340            01/01/70 00:00      
   Assembly window            01/01/70 00:00      
   hyper terminal interface            01/01/70 00:00      
      Stack size?            01/01/70 00:00      
         Stack size            01/01/70 00:00      
            Stack Management            01/01/70 00:00      
   Watchdog            01/01/70 00:00      
      stack and preview..            01/01/70 00:00      
         Limited stack            01/01/70 00:00      
      Where is the UART ISR?            01/01/70 00:00      
         ISR for UART            01/01/70 00:00      
            ISR for UART?            01/01/70 00:00      
               why "TI0=1;" is required            01/01/70 00:00      
            No, it isn't            01/01/70 00:00      
               Solved - I guess            01/01/70 00:00      
                  'polling' is NOT "standard C"            01/01/70 00:00      
                     You are right Erik..            01/01/70 00:00      
      Thought I will thank you individually..            01/01/70 00:00      
   Disable watchdog            01/01/70 00:00      
      and in Keil it is:            01/01/70 00:00      
         Status : WDT disabling in the status worked            01/01/70 00:00      

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