I'm trying to debug Dallas DS5250. All it's in a simple program but when I introduce the timer0 interrupt the debugger lost communication with the target and the interrupt is not serviced. Wich could be the problem?
int T0_ISR_CODE(int delay) __interupt #1 { TH0 = 5; TL0 = 7; } <endcode>
int T0_ISR_CODE(int delay) __interupt #1
That's not Keil C52 syntax! Besides which:
How do you expect that a parameter can be passed to an ISR?
Where do you expect that the return value for an ISR will go?
What compiler are you using? this is a Keil forum.
If, indeed, you are using Keil C51 you are making a mistake so basic that the only recourse is to work through (NOTE: not 'read') the "getting started guide", or you will only be 5 minutes before you make the next one.
anyhow a Keil interrupt should look like this void ISX_EI0 (void) interrupt 0 using 1 the 'using 1' may, of course be different.
Erik
or even omitted.