What does __attribute__((__interrupt__, no_auto_psv)) do?
void __attribute__((__interrupt__, no_auto_psv)) _T1Interrupt(void) // 5 Hz
__attribute__ directive or macro is from GCC but __interrupt__ and no_auto_psv is not , it's specific to a hardware. So, how does GCC Compiler understand __interrupt__ and no_auoto_psv, I searched and didn't find any declaration in anywhere else.
So basically the _T1Interrupt function takes no argument and return nothing but has the above attribute?
Answer
In particular, these attributes are platform-specific extensions used in the Microchip XC16 compiler for 16-bit PIC24 and dsPICs.
Attributes are essentially extra information added to the parse tree of a compiler. They exist outside the C language semantics and are there to provide additional information that the compiler uses to act consistently with your expectations. In this case __interrupt__ tells it to treat the function as an ISR (with slightly different function prolog and epilog than a normal function: dsPIC ISRs use the RETFIE return instruction, vs. RETURN for normal functions), and no_auto_psv controls whether the compiler sets the PSVPAG register:
The use of the no_auto_psv attribute omits code that will re-initialize the PSVPAG value to the default for auto psv variables (const or those placed into space auto_psv). If your code does not modify the PSVPAG register either explicitly or using the compiler managed psv or prog qualifiers then the use of no_auto_psv is safe. Also, if your interrupt service routine (or functions called by your interrupt service routine) does not use any const or space auto_psv variables, then it is safe to use no_auto_psv.