alt Hacker NewsWhy does having more more registers lead to spilling? I would assume (probably) incorrectly, that more registers means less spill. Are you talking about calls inside other calls which cause the outer scope arguments to be preemptively spilled so the inner scope data can be pre placed in registers?
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So, let's take a function with 40 alive temporaries at a point where it needs to call a helper function of, say, two arguments.
On a 16 register machine with 9 call-clobbered registers and 7 call-invariant ones (one of which is the stack pointer) we put 6 temporaries into call-invariant registers (so there are 6 spills in the prologue of this big function), another 9 into the call-clobbered registers; 2 of those 9 are the helper function's arguments, but 7 other temporaries have to be spilled to survive the call. And the rest 25 temporaries live on the stack in the first place.
If we instead take a machine with 31 registers, 19 being call-clobbered and 12 call-invariant ones (one of which is a stack pointer), we can put 11 temporaries into call-invariant registers (so there are 11 spills in the prologue of this big function), and another 19 into the call-clobbered registers; 2 of those 19 are the helper function's arguments, so 17 other temporaries have to be spilled to survive the call. And the rest of 10 temporaries live on the stack in the first place.
So, there seems to be more spilling/reloading whether you count pre-emptive spills or the on-demand-at-the-call-site spills, at least to me.
op is probably referring to the push all/pop all approach.
More registers leads to less spilling not more, unless the compiler is making some really bad choices.
Any easy way to see that is that the system with more registers can always use the same register allocation as the one with fewer, ignoring the extra registers, if that's profitable (i.e. it's not forced into using extra caller-saved registers if it doesn't want to).