x86 64 - How to interpret x86-64 instruction bytes based on the assembly instructions?

Question

I created a simple assembly program. A few Qs:

1. each instruction below starts with 0x48. However, I dont see that as a part of the instruction from this references: http://ref.x86asm.net/coder64.html#x89

E.g., x05 is the add instruction that adds immediate to a register, but not 0x0548. Similar for other instructions below.

2. How are rax, rbx encoded in the binary? E.g. Why using rbx is a byte more than rax?

     401000:   48 05 34 12 00 00       add    rax,0x1234
     40100c:   48 81 c3 34 12 00 00    add    rbx,0x1234
   

3. Same questions as bove, but a diff example. Why changing the source register to rbx, causes a change from c0 to d8?

    401006:    48 01 c0                add    rax,rax
    401009:    48 01 d8                add    rax,rbx
   

  $ objdump -d assembly_test
  
  assembly_test:     file format elf64-x86-64
  
  
  Disassembly of section .text:
  
  0000000000401000 <_start>:
    401000:   48 05 34 12 00 00       add    rax,0x1234
    401006:   48 01 c0                add    rax,rax
    401009:   48 01 d8                add    rax,rbx
    40100c:   48 81 c3 34 12 00 00    add    rbx,0x1234
    401013:   48 c7 c0 34 12 00 00    mov    rax,0x1234
    40101a:   48 89 c0                mov    rax,rax
    40101d:   48 89 d8                mov    rax,rbx

question from:https://stackoverflow.com/questions/65874396/how-to-interpret-x86-64-instruction-bytes-based-on-the-assembly-instructions

Answer 1

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