alt Hacker News> N/RVO works by (at the machine language level, of course) rewriting the function signature to return void and take an extra pointer parameter
This sounds wrong, are you sure? Would you mind demonstrating with an example on godbolt? Whether NRVO applies or not, the ABI should be the same, AFAIK.
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fluoridation • today at 1:34 PM
Yes, of that I'm sure. This optimization is only possible if the compiler has control of both sides of a call. If the function may be callable from other translation units or modules I imagine it generates a thin wrapper that's externally callable.
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Yes, it works exactly like this, this is a demo on godbolt [0]. rdi stores the pointer in both cases, makeS1() uses RVO, makeS2() takes it explicitly and constructs with placement new.
I will say before testing this i didn't realize the RVO calling convention was to return the pointer you pass in, but apparently so. If makeS2() returned void, it's just a tail call to the constructor, but makeS1() has to spill rbx and use it to save the pointer.
[0]: https://godbolt.org/z/ovd1n99P8