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Many websites and textbooks state that CISC instructions are easier or faster to compile than RISC instructions. I've been having a hard time finding a definite, comprehensive reason for this, one textbook states that it's because CISC instructions resemble high-level code more closely but I'm still not sure why that makes them easier/faster to compile. Does compiling CISC instructions take less time simply because there's less of them in a program or is there something else I'm missing?

Peter Cordes
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Pzet
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    It's just easier from a human POV, or if doing fully naive code-gen without optimization so all variables are kept in memory, not registers. Having more options (like addressing modes, and memory source vs. register) to choose from when doing CISC code gen means more possibilities to consider when optimizing. – Peter Cordes Aug 10 '20 at 20:35
  • So the extra features that come with CISC make it so the assembly code can be adapted to different scenarios during compilation? – Pzet Aug 10 '20 at 20:43
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    You've gotr hold of the wrong ed of the stick here. You don't compile machine instructions, you assemble them, or generate them. CISC is easier for the compiler as a *target* language,to *generate*, because it is richer, and so you don't have to generate so much code or worry about registers so much. – user207421 Aug 10 '20 at 21:46

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