Hi all:
I am confused by the interrupt vector table of linux in arm920t. When an interrupt arrives, pc SHOULD load the address of the interrupt handler, it works smoothly with MMU DISABLED. When with MMU ENABLED, how does PC be loaded the address of the interrupt handler,eg,
0xc0002000? Firstly, I think pc = 0x0, and map the 0x0(virtual address) to an physical address. But it seems that no mapping table exists. Since the interrupt handler belongs to kernel, it must be greater than 0xc000_0000, and 0x0 must belong to user space(MMU enabled.) In intel x86 cpu, IT HAS A IDTR to indicate the address WHICH ip LOADS according to the IDT, which is the ARM920T if it exists?
Please give me an explanation. Thanks in advance.
Michael