What kind of interrupt arrangement is this for msp430f5438?

水货老手

What kind of interrupt arrangement is this for msp430f5438? [Copy link]

// Timer1 A0 interrupt service routine #pragma vector=TIMER1_A0_VECTOR __interrupt void Timer_A0 (void) { TA1CCR0 += 4; // Add Offset to CCR0 } // Timer_A3 Interrupt Vector (TAIV) handler #pragma vector=TIMER1_A1_VECTOR __interrupt void TIMER1_A1_ISR(void ) { switch(__even_in_range(TA1IV,14)) { case 0: break; case 2: TA1CCR1 += 16; // Add Offset to CCR1 break; case 4: TA1CCR2 += 100; // Add Offset to CCR2 break; case 6: break; // CCR3 not used case 8: break; // CCR4 not used case 10: break; // CCR5 not used case 12: break; // Reserved not used case 14: P4OUT ^= 0x10; // overflow break; default: break; } } Please help me, why does it separate CCR0? What is the logic? I just learned 5438 and don't understand it very well.

gongshuipeng

Because #pragma vector=TIMER1_A0_VECTOR uses a single interrupt vector entry, which has a higher priority than the others, while the others share one entry. Read more books. [ This post was last edited by gongshuipeng on 2011-7-20 14:13 ]

水货老手

Yes

whgwuhan

The author of the post is so sharp. It took me many days to put together this program. I admire you for analyzing it to this point!

#pragma vector=TIMER1_A1_VECTOR
__interrupt void TIMER1_A1_ISR(void)
{
switch(__even_in_range(TA1IV,14))
{
case 0: break;
case 2: TA1CCR1 += 16; // Add Offset to CCR1
break;
case 4: TA1CCR2 += 100; // Add Offset to CCR2
break;
case 6: break; // CCR3 not used
case 8: break; // CCR4 not used
case 10: break; // CCR5 not used
case 12: break; // Reserved not used
case 14: P4OUT ^= 0x10; // overflow
break;
default: break;
}
}