Dear Champs,
I am asking this for our customer.
In 6.9.15 The INTERRUPT Pragma of the compiler doc (https://www.ti.com/lit/ug/spru514w/spru514w.pdf?ts=1630324430532&ref_url=https%253A%252F%252Fwww.ti.com%252Ftool%252FC2000-CGT), I want clarify usages.
1) For interrupt C28x ISR and CLA task,
If #pragma INTERRUPT is used, does it matter with or without interrupt?
That is, are below a) b) c) the same?
a) interrupt void Cla1Task1()
b)
#pragma INTERRUPT (Cla1Task1)
interrupt void Cla1Task1()
c)
#pragma INTERRUPT (Cla1Task1)
void Cla1Task1()
2) Likewise, for CLA background, are below the same?
a)
#pragma INTERRUPT(Cla1Task8_background, "BACKGROUND");
interrupt void Cla1Task8_background()
b)
#pragma INTERRUPT(Cla1Task8_background, "BACKGROUND");
void Cla1Task8_background()
I tried to build both the above and did not find anything different, but I still want to confirm again.
3) In the example shown on the page 123, there is a ";" on the end of #pragma INTERRUPT. Does it matter if ";" is used or not? I don't see a ";" with other #pragma usages.
#pragma INTERRUPT(task_name, "BACKGROUND");
void task_name(void) {...}
4) If I define
#pragma INTERRUPT(Cla1Task8_background, "BACKGROUND");
interrupt void Cla1Task8_background()
in the .h file, I only need to use extern without #pragma statement, is it right?
extern interrupt void Cla1Task8_background();
Wayne Huang
