Question

I was looking in to some compiler generated assembly code(x86_64) and found few instructions which have a zero displacement as following..

     movzbl 0x0(%rbp),%eax

What could be the reason behind such 0x0 displacement? (I am new to assembly, please point me to the links if its already a discussed issue.) From my understanding, the above instruction copies the zero extended rbp+(0x0) to eax.

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EDIT: I found a link which explains it for leal though
what-does-0x0-indicate-in-the-instruction

Thank you..!

Answer 1

It means movzx eax, byte [rbp + 0].

There’s no encoding for memory operands at address contained in (e|r)bp, but there is for (e|r)bp + constant. So, you have 0 for that constant to extract a byte from the address in rbp.