Question

This section has me thoroughly confused. I have an example problem that I am hoping someone can break down into steps for me so that I may absorb how it is done to apply to other problems.

mc:   call subr
mr:   mov  [val],ax
subr: push ax
      push bx
      push cx
      add  ax,dx
      pop  ax
      pop bx
      pop cx
      ret

The book asks what the hex value in the sp and ax register will be for when the code returns from the subroutine and reaches the instruction mr: mov [val],ax. sp=0100 ax=0002, but I have no clue as to how to arrive at these answers.

The instruction mc: call subr saves the address of the next sequential instruction mr: mov [val],ax on the stack so the subroutine can correctly return. The absolute address in memory where the return address is saved is 1120E. Can anyone please elaborate on this as well?

registers given:
ax = 0000  bx = 0001  cx = 0002  dx = 0004
si = 0000  di = FFFF  bp = 0080  sp = 0100
cs = 1000  ds = 1100  es = 1110  ss = 1111

Answer 1

The call at mc: will save the current address, so when the subr: returns, control will start again at mr:.

Since subr: pushes ax, bx, and cx in that order. It then pops ax, bx, and cx in that order, so what was pushed from cx gets popped into ax (and vice versa). The effect of these is to swap ax and cx. The add ax, dx has no real effect on the result produced, because immediately after adding dx to ax, it pops ax off the stack. The add does affect the flags, but nothing here does anything based on the flags, so at least in the code you’ve shown, that doesn’t mean much either.

After control returns to mr:, it writes the value in ax to memory, and then flows back into subr:, thus swapping ax and cx back to where they started.

IOW, the whole this is a very slow, roundabout way of achieving roughly the same effect as:

mc: mov [val], cx
    ret

As far as the absolute address goes, there’s not much to say. In particular, chances are pretty good that if you run the same code again, it may be loaded at a different address, o the address saved on the stack may be entirely different.