Why does a function start with push %rbx + mov %rdi, %rbx? What does this two-instruction pattern mean?

It stashes the first argument (%rdi) into a callee-saved register (%rbx) so the value survives the upcoming calls that would otherwise clobber %rdi.

* The arg in caller-saved %rdi is stashed into callee-saved %rbx so it survives a later call that clobbers %rdi. *

phase_1:
    push %rbx
    mov  %rdi, %rbx
    call <phase_init>
    mov  $0x402490, %esi
    mov  %rbx, %rdi
    call <strings_not_equal>

Step by step:

1. push %rbx — saves the old %rbx (callee-saved, we must preserve it)
2. mov %rdi, %rbx — copies our input string pointer to %rbx
3. call <phase_init> — this call may trash %rdi (it's caller-saved!) but %rbx is safe
4. mov %rbx, %rdi — restore the argument from %rbx back into %rdi for the next call

The key insight: %rdi is caller-saved — any call can destroy it. If the function needs the argument again later, it must stash it somewhere safe. A callee-saved register (%rbx, %rbp, %r12-%r15) is guaranteed to survive across calls.

General rule: If you see push %r12 + mov %rdi, %r12 at function entry, it means "I need the first argument to survive across function calls."

Go deeper:

• System V AMD64 calling convention — why %rdi is caller-saved and %rbx/%r12-r15 survive across calls