Do C++11 compilers turn local variables into rvalues when they can during code optimization?

Question

Sometimes it\'s wise to split complicated or long expressions into multiple steps, for example (the 2nd version isn\'t more clear, but it\'s just an example):

Answer 1

Be aware that besides move semantics that can greately speed up your code, compiler is also doing (N)RVO - (Named) Return Value Optimization, which can actually give even more efficiency to your code. I have tested your example and in g++4.8 it appears that your second example could be actually more optimal:

object3 a(x);
object2 b(a);
object1 c(b);
return c;

From my experiments it looks like it would call constructor/destructor 8 times (1 ctr + 2 copy ctrs + 1 move ctr + 4 dtrs), compared to other method that call it 10 times (1 ctr + 4 move ctors + 5 dtors). But as user2079303 has commented, move constructors should still outperform copy constructors, also in this example all calls will be inlined so no function call overhead would take place.

Copy/move elision is actually an exception to "as-if" rule, that means that sometimes you may be suprised that your constructor/destructor even tho with side effects does not get called.

http://coliru.stacked-crooked.com/a/1ca7ebec0567e48f

(you can disable (N)RVO with -fno-elide-constructors parameter)

#include 
#include 

template
struct A {
    A() { std::cout<<"A::A"<
    A(const A&) { std::cout<<"A::A&"<
    A(const A&&) { std::cout<<"A::A&&"< foo () {    
    A<2> a; A<1> b(a); A<0> c(b); return c;   // calls dtor/ctor 8 times
    //return A<0>(A<1>(A<2>()));  // calls dtor/ctor 10 times
}
int main()
{
   A<0> a=foo();
   return 0;
}