x86指令addq 16(%rbp),%rax转y86等效代码验证问询

Hey there! Let's break this down step by step for you.

Your interpretation of addq 16(%rbp), %rax is 100% correct! Here's the exact breakdown of how it executes:

• First, calculate the effective address: %rbp + 16 (note: the 16 here is a byte offset, since x86-64 uses byte addressing for memory)
• Read a 64-bit (qword) value from that calculated memory address
• Add this fetched value to the current contents of %rax
• Store the resulting sum back into %rax

Your provided Y86 code starts with:

rrmovq %rsp, %rbx
iaddq $0x16, %rbx
m...

Let's clarify its functionality based on what's written and what it likely needs to be:

• As written so far, it uses %rsp as the base register instead of %rbp—this would not match the original x86 instruction, since the x86 code uses %rbp to compute the memory address.
• If this is a typo, and you intended to use %rbp instead of %rsp, the full correct Y86 implementation would look like this:

  rrmovq %rbp, %rbx   ; Copy %rbp's value to %rbx
  iaddq $0x16, %rbx   ; Compute %rbp + 16, store in %rbx
  mrmovq (%rbx), %rcx ; Load the 64-bit value from %rbx's address into %rcx
  addq %rcx, %rax     ; Add %rcx to %rax, store result in %rax
  

  This complete snippet would perfectly replicate the behavior of the original x86 instruction. A quick note: Y86 doesn't support memory operands directly in arithmetic instructions (unlike x86's addq mem, reg syntax), so you have to split the operation into loading the memory value to a register first, then performing the addition.

内容的提问来源于stack exchange，提问作者Shinji-san