问题
Trying to understand the cause of the watchdog. The snippet of assembly routine is as follows:
fe813cf0: 2820 cmp r0, #32
fe813cf2: dbf0 blt.n fe813cd6 <XYZ+0x10>
fe813cf4: f04f 30ff mov.w r0, #4294967295 ; 0xffffffff
fe813cf8: bd10 pop {r4, pc}
My PC value from dump is fe813cf3. What does this mean? How should I interpret it? Please let me know if more info is required.
回答1:
If it's a watchdog that triggers, that seems to indicate that the processor has hung. The state you're capturing might be corrupt, in the worst case. Not sure how you get the "dump", so it's hard to tell.
In general on ARM, an odd address indicates a jump that wanted to switch to the Thumb instruction set.
See for instance this documentation of the BX (branch exchange) instruction:
The BX instruction branches to the address contained in a specified register. The value of bit 0 of the branch address determines whether execution continues in ARM state or Thumb state.
Bit 0 of an address can be used in this way because:
All ARM instructions are word-aligned. This means that bits 0 and 1 of the address of any ARM instruction are ignored because these bits refer to the halfword and byte part of the address.
All Thumb instructions are halfword-aligned. This means that bit 0 of the address of any Thumb instruction is ignored because it refers to the byte part of the address.
However, I'm pretty sure the above implies that the PC is never actually set to an odd address, bit 0 is cleared by the branch instruction.
来源：https://stackoverflow.com/questions/16079189/pc-value-in-between-two-instructions-program-counter