Why do x86-64 instructions on 32-bit registers zero the upper part of the full 64-bit register?

Question

In the x86-64 Tour of Intel Manuals, I read

Perhaps the most surprising fact is that an instruction such as MOV EAX, EBX automatically ze

Answer 1

I'm not AMD or speaking for them, but I would have done it the same way. Because zeroing the high half doesn't create a dependency on the previous value, that the CPU would have to wait on. The register renaming mechanism would essentially be defeated if it wasn't done that way.

This way you can write fast code using 32-bit values in 64-bit mode without having to explicitly break dependencies all the time. Without this behaviour, every single 32-bit instruction in 64-bit mode would have to wait on something that happened before, even though that high part would almost never be used. (Making int 64-bit would waste cache footprint and memory bandwidth; x86-64 most efficiently supports 32 and 64-bit operand sizes)

The behaviour for 8 and 16-bit operand sizes is the strange one. The dependency madness is one of the reasons that 16-bit instructions are avoided now. x86-64 inherited this from 8086 for 8-bit and 386 for 16-bit, and decided to have 8 and 16-bit registers work the same way in 64-bit mode as they do in 32-bit mode.

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See also Why doesn't GCC use partial registers? for practical details of how writes to 8 and 16-bit partial registers (and subsequent reads of the full register) are handled by real CPUs.