Two Saves One Switch

A timer interrupt arrives while process A is running, and the OS decides to switch to process B. Here is the sequence:

#	Who acts	Action
1	hardware	timer interrupt fires
2	hardware	save registers, move to kernel mode, jump to the trap handler
3	OS	handle the trap, decide to switch, call `switch()`
4	OS	save registers, restore B's, change the stack pointer to B's kernel stack
5	OS	execute return-from-trap
6	hardware	restore B's registers, move to user mode, jump to B's PC

Steps 2 and 4 both save registers. Which describes them correctly?

Step 2 saves into A's process structure and step 4 onto A's kernel stack, so the deeper save happens closer to the switch

Step 2 saves A's user registers onto A's kernel stack, implicitly, in hardware. Step 4 saves A's kernel registers into A's process structure, explicitly, in software

Both save to A's kernel stack. Step 2 saves the general-purpose registers and step 4 saves the program counter and stack pointer

Both save into A's process structure. Step 2 is the fast path and step 4 only runs when the OS actually switches processes