arminterrupt-handlingarmv7irq

How to calculate the LR value when ARMv7-A architecture implementation takes an IRQ exception

I'm researching the the Arm Architecture Reference Manual ARMv7-A ARMv7-R edition document these days. When I read about the exception handling part of this manual, it comes across a confusions to me. The problem is about how to decide the LR value when ARMv7-A architecture implementation takes an IRQ exception.

EXAMPLE: Suppose that the processor is executing an instruction at the address of 0x_0000_1000 and an IRQ is taken.First, we have to calculate some parameters to be used to calculate the LR.

  1. preferred return address,which is the Address of next instruction to execute in this case. So preferred return address = 0x_0000_1002 in thumb instruction set or preferred return address = 0x_0000_1004 for arm instruction set.preferred return address for the exception
  2. PC,which is the program counter and holds the current program address.In this case, PC = 0x_0000_1004 in thumb instruction state or PC = 0x_0000_1008 in arm instruction state.how to calculate PC

Then, here are 2 methods mentioned in the document to decide the LR value when taking this IRQ exception.

  1. by using preferred return address. LR = preferred return address + offset that depends on the instruction set state when the exception was taken.In this case LR = 0x_0000_1002 + 4 in thumb instruction state or LR = 0x_0000_1004 + 4 in arm instruction state.Offsets applied to Link value for exceptions taken to PL1 modes
  2. by using PC. LR = PC-0 if in thumb instruction set or LR = PC-4 when in arm instruction set.In this case LR = 0x_0000_1004 - 0 in thumb instruction set or LR = 0x_0000_1008 - 4 in arm instruction state. Pseudocode description of taking the IRQ exception

Problem:the LR results calculated by the 2 methods are different both in thumb set state and arm set state(with first method we get LR = 0x_0000_1006 or LR = 0x_0000_1008,but second method we get LR = 0x_0000_1004 or LR = 0x_0000_1004). which one is correct or is there any wrong with my understanding?

Solution

  • TL;DR - the IRQ LR will point to the next instruction to complete work as would normally be run without an interrupt. Otherwise, code would not execute the same in the presence of interrupts.

    It is confusing as the ARM documents may refer to PC in many different contexts and they are not the same.

    EXAMPLE:Suppose that the processor is executing an instruction at the address of 0x_0000_1000 and an IRQ is taken. First, we have to calculate some parameters to be used to calculate the LR.

    preferred return address,which is the Address of next instruction to execute in this case. So preferred return address = 0x_0000_1002 in thumb instruction set or preferred return address = 0x_0000_1004 for arm instruction set.preferred return address for the exception

    This is not correct. The ARM cpu has a pipeline and the last instruction that it has deemed to have completed is the next instruction. Take for example this sequence,

         0: cmp r1, #42
     1: bne 4f          ; interrupt happens as this completes.
     2: add r2, r2, #4
     3: b   5f
     4: sub r2, r2, #2
     5: ; more code.

    If the interrupt happens as label '1:' happens, the next instruction will be either '2:' or '4:'. If you followed your rule this would either increase interrupt latency by never allowing an interrupt in such cases, or interrupts would cause incorrect code. Specifically, your link says next instruction to execute.

    PC,which is the program counter and holds the current program address.In this case, PC = 0x_0000_1004 in thumb instruction state or PC = 0x_0000_1008 in arm instruction state.how to calculate PC

    Here you are mixing concepts. One is when you use a value like ldr r0, [pc, #42]. When you calculate the offset, you must add two to the current ldr instruction. The actual PC is not necessarily this value. At some point (original version), the ARM was a two stage pipeline. In order to keep behaviour the same, subsequent ARM cpus follow the rule of being two ahead when calculating ldr r0, [pc, #42] type addresses. However, the actual PC may be much different inside the CPU. The concept above describes the programmer visible PC for use with addressing.

    The CPU will make a decision, sometimes base on configuration, on what work to complete. For instance, ldm sp!, {r0-r12} may take some time to complete. The CPU may decide to abort this instruction to keep interrupt latency low. Alternatively, it may perform 12 memory reads which could have wait states. The LR_irq will be set to the ldm instruction or the next instruction depending whether it is aborted or not.