--- At Thu, 24 Apr 2003 16:56:40 +0100, Ben Hutchings wrote:

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dick.bridges@tais.com wrote:

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I don't understand C++ well enough to grok a compiler's interpretation of the code, but I seems like I'm creating an extra copy for the return. <snip>

Logically the local variable is copied to the the caller, but since

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of the returned expression matches the return type the compiler is allowed to optimise away that copy, even though the return type has a user-defined copy constructor. In this case you always return a certain named variable, so the compiler can arrange to construct it in the memory location where the caller expects the returned value. This is called the named-return-value optimisation, or NRVO.

I dont want to start any kind of war, and this is a bit off topic, but I was investigating NVRO and found a few discussions with Scott Meyers

Duane Murphy wrote: the type that

indicated that naming is not relevant to the optimization. That this:

shared_ptr<int> maybe_get_a_pointer() { return shared_ptr<int>(); }

is perfectly reasonable and can also be optimized.

Is there more recent and complete discussion concerning this optimization? Is naming actually required?

I realise that the implementation is allowed to optimise out the copy whether the return value is named or not. Actually doing so requires quite different treatment of the two cases, so it's worth considering them as separate optimisations. For NRVO to be possible at some return statement, there must not be any return statements reachable after the construction of the returned object that return any other object. The optimisation involves changing the code generated for the definition of the object, rather than the code for the return statement. So the implementation has to consider the whole function when doing this. Anonymous RVO is a very local optimisation, so it should require less work to apply.