x86 - Not sure why we add the registers %rdx and %rax when the assembly code has been using %eax and %edx

Question

Hello I need some help understanding what is going on in this assembly code:

        .file   "mystery.c"

        .text

        .globl mystery

              .type mystery, @function

 mystery:
   pushq    %rbp
    movq    %rsp, %rbp

   movl %edi, -20(%rbp)
   movl $1, -16(%rbp)
   movl $0, -12(%rbp)
   movl $0, -8(%rbp)
   cmpl $2, -20(%rbp)
   jg   .L2
   movl $1, %eax
   jmp  .L3

  .L2:
movl    $2, -4(%rbp)
jmp .L4

  .L5:
movl    -12(%rbp), %eax
movl    -16(%rbp), %edx
leal    (%rdx,%rax), %eax
movl    %eax, -8(%rbp)
movl    -16(%rbp), %eax
movl    %eax, -12(%rbp)
movl    -8(%rbp), %eax
movl    %eax, -16(%rbp)
addl    $1, -4(%rbp)

.L4:
movl    -4(%rbp), %eax
cmpl    -20(%rbp), %eax
jle .L5
movl    -8(%rbp), %eax

.L3:
leave
ret

I understand exactly what is going on UNTIL I get to .L5, Here the command leal(%rdx, %rax), eax is what is confusing me. Up until now ive been moving values to eax and edx and now im adding the values in rdx and rax. Where is rdx and rax coming from and what values are they holding? Are they just another way of writing eax and edx? Thanks for any help.

Answer 1

See this related answer. It explains the different registers and their evolution. In this case, the %rax register is a 64 bit register. %eax is the 32 bit one, and %ax would be 16 bits. %ah refers to the high 8 bits of the 16 bits in the register, and %al refers to the lower end.

This little diagram was taken from another answer to the same question, but it shows it well...

|63..32|31..16|15-8|7-0|
               |AH.|AL.|
               |AX.....|
       |EAX............|
|RAX...................|