汇编编程：如何将sumsquare方法结果传回栈？代码修改咨询

Great question! Let's walk through how to modify your sumsquare routine to return the result directly to the stack instead of using the RR register.

First, let's recap the original stack setup when calling sumsquare:

• Before bsr sumsquare, your stack holds the parameters a and b (structure: [..., a, b]).
• Inside sumsquare, the link 1 instruction creates a stack frame with 1 local variable, making the parameters accessible at offsets -3 (for a) and -2 (for b) relative to the frame pointer.

Your goal is to replace the str RR instruction (which stores the result to the RR register) with an instruction that writes the result directly to the stack, overwriting the original parameters. Here's the solution:

Modified sumsquare Routine #

sumsquare:
link 1 // 1 local variable
ldl -3 // a
ldl -3 // a
mul // a*a
ldl -2 // b
ldl -2 // b
mul // b*b
add // a*a + b*b → result is on top of stack
// Replace str RR with this line:
stl -3 // Store the result directly into the stack position of parameter 'a'
unlink 1
ajs -1 // Remove the now-unneeded 'b' parameter from the stack
ret

Key Changes Explained: #

1. Replace str RR with stl -3:

   • The stl instruction stores the value on top of the stack to a stack location relative to the frame pointer.
   • Using -3 targets the original stack position of parameter a, overwriting it with our computed result.

2. Add ajs -1 before ret:

   • After unlinking the stack frame, our stack still contains the original b parameter. The ajs -1 adjusts the stack pointer to remove this extra entry, leaving only our result on the stack when the routine returns.

Simplified Caller Code #

You can streamline the caller since we no longer need to delete parameters or load from RR:

ldc 5 // a
ldc 12 // b
bsr sumsquare
stl 1 // Directly store the stack-top result into your variable
....

That's it! This setup makes sumsquare return the result directly to the stack, eliminating the need for the result register.

内容的提问来源于stack exchange，提问作者John Hm