What is register %eiz?
assembly, gnu-assembler, objdump, x86
Solution
See Why Does GCC LEA EIZ?:
Apparently `%eiz` is a pseudo-register that just evaluates to zero at all times (like `r0` on MIPS).
...
I eventually found a mailing list post by binutils guru Ian Lance Taylor that reveals the answer. Sometimes GCC inserts NOP instructions into the code stream to ensure proper alignment and stuff like that. The NOP instruction takes one byte, so you would think that you could just add as many as needed. But according to Ian Lance Taylor, it’s faster for the chip to execute one long instruction than many short instructions. So rather than inserting seven NOP instructions, they instead use one bizarro LEA, which uses up seven bytes and is semantically equivalent to a NOP.
Problem
In the following assembly code that I dumped out using `objdump`: ``` lea 0x0(%esi,%eiz,1),%esi ``` What is register `%eiz`? What does the preceding code mean?