??? 12/01/08 18:36 Read: times |
#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 |
Topic | Author | Date |
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 |