You’re missing the fact that the compiler isn’t forced to fill every register in the first place. If it was less efficient to use more registers, the compiler simply wouldn’t use more registers.

The actual counter proof here would be that in either case, the temporaries have to end up on the stack at some point anyways, so you’d need to look at the total number of loads/stores in the proximity of the call site in general.

This argument doesn’t make sense to me. Generally speaking, having more registers does not result in more spilling, it results in less spilling. Obviously, if you have 100 registers here, there’s no spilling at all. And think through what happens in your example with a 4 register machine or a 1 register machine, all values must spill. You can demonstrate the general principle yourself by limiting the number of registers and then increasing it using the ffixed-reg flags. In CUDA you can set your register count and basically watch the number of spills go up by one every time you take away a register and go down by one every time you add a register.

I recalled there were some new instructions added that greatly help with this. Unfortunately I'm not finding any good _webpages_ that describe the operation generally to give me a good overview / refresher. Everything seems to either directly quote published PDF documents or otherwise not actually present the information in it's effective for end use form. E.G. https://www.felixcloutier.com/x86/ -- However availability is problematic for even slightly older silicon https://en.wikipedia.org/wiki/X86-64

- XSAVE / XRSTOR

- XSAVEOPT / XRSTOR

- XSAVEC / XRSTOR

- XSAVES / XRSTORS

A good compiler will only do that if the register spilling is more efficient than using more stack varibles, so I don't really see the problem.