How to declare an interrupt handler (ISR) in Mspgcc?

msp430 mspgcc

Wayne Uroda · Mar 19, 2013 · Viewed 7.7k times · Source

What is the preferred method of declaring an interrupt handler in mspgcc?

Answer

Note that this information applies to MSPGCC v4.6.3 which pre-dates the Ti/Redhat/SOMNIUM port of MSPGCC which is now distributed with code composer studio.

Thanks go to Peter Bigot of the Mspgcc-users mailing list for this answer.

There are two possibilities:

A. Use Code composer studio style syntax (has the added benefit of being portable to CCS):

#pragma vector=TIMER0_A0_VECTOR    
__interrupt void
ta0cc0_isr (void)

(Note that this was introduced somewhere around version 20120406 of Mspgcc).

B. Use native gcc syntax:

static void
__attribute__((__interrupt__(TIMER0_A0_VECTOR)))
isr_cc0_TA0 (void)

C. Name the function correctly so that it is included into the vector table (useful for ASM functions).

The interrupt attribute causes the function to be named __isr_X where X is the word offset of the interrupt from the vector table start (equal to the value of the interrupt attribute's parameter divided by 2).

These __isr_X symbols are used to initialize the vector table in crt0ivtbl.o.

How to declare an interrupt handler (ISR) in Mspgcc?

Answer

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