Confusion on push/pop

[unktehi]

In my class I have the following example that represents the equation:
eax = (ecx * edx) / eax

These are the lines that they wrote to show how this is done:

push edx
(save edx for later)

push eax
(save eax for later)

mov eax, ecx
(save ecx in eax: eax = ecx)

imul edx
(multiply eax (now ecx) x edx)

pop ebx
(where did ebx come from? The last value that was pushed was eax, so if the instruction uses a different register does it assume the last thing that was pushed? Is ebx popping the value of what was formerly pushed as eax?)

idiv ebx
(This is dividing edx: eax (from 2 steps above) with ebx (which was the former value of eax before it was pushed?)

pop edx
Restore the edx value to the original 'pushed' value

Do I understand this correctly according to the comments under each instruction?

[Farabi]

Yes ebx is the value from the last push and it is eax.
Try to see it on debugger to understand it better.

[cmpxchg]

imul edx
(multiply eax (now ecx) x edx)

imul edx
multiply eax by edx (eax x edx), ecx is not even touched and not changed

push edx
push eax
pop ebx       ebx = eax
pop edx       

the only requirement is to have  equal # of push & pop commands, 2 push & 2 pop (because push/pop modify esp register)
push/pop can be used to transfer value from one register to another, this is exactly what you code does

push eax - saves contet of eax in memory then any instruction can be used to read memory(no more dealing with eax)
pop  ebx - reads memory and transfers content to ebx
reason how pop knows where to read is esp register - your pointer to memory, actively used but the use is hidden

in you case here an alternative to 'pop ebx' would be:
mov  ebx, dword [esp]
add  esp, 4

[unktehi]

Thanks for confirming my understanding.  I appreciate it!