Why the program don't enter in this intrrerupt??
this is my code
void timer0_isr (void) interrupt 1 using 1 { int a=0,c; a++; // ram[a] = din(7); c=9; if (a > 10) a=0; }
If your program never enters the interrupt service routine, how are we supposed to tell that from the code of the routine ?
i write that value for c and i see the program canot execute this
If the processor never enters the interrupt routine, then the error IS NOT inte the interrupt service routine, hence not in the code you have posted.
Yes, your ISR can be buggy, but the reason why it isn't run is somewhere else!
the code is this
// Set Timer0 for the fix rate 1464.8 Hz = 12MHz / 2^13 // TL0 = 0x27; // TH0 = 0xF1; TMOD = 1; ET0 = ON; // Timer 0 Interrupt on TR0 = ON; // Timer 0 start
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