problem of interrupt vector table of linux in arm920t

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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

Please give me an explanation. Thanks in advance.


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