I'm not exactly sure what this x86 Add instruction is doing

I'm not exactly sure what this add instruction is doing:

add 0x0(%rbp,%rbx,4),%eax

If it were:

add %rbx,%eax

I know it would add the contents of rbx and the contents in eax and store them back into eax. However, t开发者_高级运维he 0x0(%rbp,%rbx,4) is throwing me off.

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That's because it's stupid&confusing AT&T syntax.
In normal Intel syntax it's add eax,dword ptr[rbp+4*rbx+0] ie add the dword at rbp+4*rbx to eax.