<aside> <img src="/icons/map-pin_gray.svg" alt="/icons/map-pin_gray.svg" width="40px" /> Relevant Notes/Resources:

4 Advanced Assembly Programming

Instruction Set

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• Question

Enter the Value

The concept is pretty similar to what we did for the midterm. The entirety of the question usually revolves around the instruction swi, which performs the following operation:

As noted in the operation, SP refers to the stack pointer. But, also as a reminder, a stack goes from a high address to low address. So when we subtract, we go “up” the table.

In general, the stack after swi is executed should look something like this:

I’ll provide a brief explanation on the first two operations, as the rest pretty much follow the same concept:

• $(SP) - 2 \\implies SP;$

  The stack pointer goes up the table, since $2500 - 2 = $24FE.

  

• $RTN_H:RTN_L \\implies M_{(SP)}:M_{(SP+1)}$;

  The following means to store $RTN_H$ in $M_{(SP)}$ and $RTN_L$ in $M_{(SP+1)}$. So how does that exactly translate to the stack content:

  

  • $M_{(SP)}$ is the memory of the stack pointer is at, which if you recall, was shifted up by 2.
  • $M_{(SP+1)}$ is the memory below the stack pointer, because again as a reminder, the stack goes from high to low address. So by adding, you move down a row.

Then, this repeats for $Y_H:Y_L$ , $X_H:X_L$ , $B:A$ and $CCR$. At which, you should obtain the following stack frame:

The solution for the other question is:

1. ldd #$7843 load register D with $7843.
   • As such, accumulator A is $78 and accumulator B is $43.
2. aba adds accumulator A and B, and loads the results to accumulator A.
   • The hexadecimal addition results in $78 + $43 = $BB.
   • As such, register D is now $BB43, since accumulator A is $BB and accumulator B remains the same, which is $43.