According to ARM Architecture Procedure Call Standard (AAPCS) on the ARMv6-M, and ARMv7-M architecture in it says:
"Although the processor hardware allows SP to be at any word aligned address at function boundaries, standard programming practice requires C program code to ensure that the SP is at a 64-bit (doubleword) aligned address."
What does it mean that the Stack pointer has to be at a 64 bit aligned address?
For some legacy processors the LDRD in ARM state has 8 byte alignment requirement. But no such requirement for LDRD in ARMv7-M.
regards,
Joseph
Thanks for that. As they say it's the things you know but ain't so that catch you.
Yes, indeed, this is helpful and good information.
-So there will be no problems with alignment faults regarding LDRD if SP is not aligned on an 8-byte boundary.
I would expect that keeping the 8-byte boundary might help when it comes to cache and speed.
Yes. For processors with 64-bit interface (e.g. Cortex-M7), having such data alignment would be more efficient. It could also made debugging easier (e.g. when setting up data watchpoint).
It also avoids corner cases where a 64-bit access go across boundaries of different memory types, which can result in unpredictable behavior according the the architecture specification.