2024/09/23

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2024-09-23 16:39:06 +0200 <raehik> I figure I could benchmark the instructions themselves with C if I wanted, but I assume there's some programmer knowledge out there like "your average ALU will do all these operations at the same speed"
2024-09-23 16:37:39 +0200 <raehik> oh. I was most interested in instruction count. I figured that would tell me enough
2024-09-23 16:36:44 +0200 <merijn> Like, we're talking performance differences that would probably take you weeks of micro-benchmark writing to determine
2024-09-23 16:36:41 +0200 <raehik> merijn: that's true, but I figured I could google for general case comparing XOR with whatever gets used for `eqX`
2024-09-23 16:36:01 +0200 <merijn> Less instructions is not the same as faster
2024-09-23 16:35:44 +0200 <merijn> I mean, that doesn't tell you much either
2024-09-23 16:35:12 +0200 <raehik> atrociously pointless optimization for the error case which should not happen much xd
2024-09-23 16:33:47 +0200rosco(~rosco@175.136.158.234) (Quit: Lost terminal)
2024-09-23 16:32:51 +0200 <raehik> merijn: ok thx just going by feel here. think I do both and compare output assembly
2024-09-23 16:31:36 +0200 <raehik> (actually probably I want xor not and but w/e)
2024-09-23 16:31:30 +0200 <raehik> In my head, I can make up some assembly where both are the same number of instructions (assuming 8-byte instructions). But it would rely on doing a conditional jump on the boolean op result, which I don't know how to control
2024-09-23 16:31:26 +0200 <merijn> raehik: Probably not in any sensible manner :p
2024-09-23 16:29:29 +0200 <raehik> I want to check two Word64# s for equality, but in the event of inequality, I want to see which byte failed first (from left to right). I could do this with `and64#` and checking the output. Would this be any slower than using `eqWord64#`?
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