As follow the Reference manual show,
Fast Interrupts vector directly to a service routine based on values in the Fast Interrupt Vector Address registers without having to go to a jump table first.
Below is my Timer 2 initial, I had designated the interrupt subroutine to register.
So you can see the SystemTickCount value has increased in the main void. It mean it can enter interrupt.
And the address 0x0000 to 0xDF is normal.
The following is my question.
This is my memory planning, the Vector table is at address 0x0000, and my program is start at 0x4208.
When I erase the address from 0x0000 to 0x800 (one sector ) As below picture show.
You can see that is a problem with the execution of the following program.
The CPU thread jump to an unknown address and cannot execute a normal program.
According to the instructions of register, it should jump directly to the subprogram without having to go to a jump table first. Right?
Am I missing something?
For the Fast interrupt mechanism, the interrupt vector and corresponding are saved in INTC_FIMx and INTC_FIVALx/VAHx, so you are not required to write the Fast Interrupt vector in the vector table.
Regarding your question that the CPU thread jumps to an unknown address and cannot execute a normal program, I think it is not related to fast interrupt, there are the other cause.
BTW, can you tell us the part number you are using?