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):

return object1(object2(object3(x))); 

can be written as:

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

Assuming all 3 classes implement constructors that take rvalue as a parameter, the first version might be faster, because temporary objects are passed and can be moved. I'm assuming that in the 2nd version, the local variables are considered to be lvalues. But if the variables aren't later used, do C++11 compilers optimize the code so the variables are considered to be rvalues and both versions work exactly the same? I'm mostly interested in Visual Studio 2013's C++ compiler, but I'm also happy know how the GCC compiler behaves in this matter.

Thanks, Michal

Answer 1

The compiler cannot break the "as-if" rule in this case. But you can use std::move to achieve the desired effect:

object3 a(x); object2 b(std::move(a)); object1 c(std::move(b)); return c; 

Answer 2

As juanchopanza said, the compiler cannot (at C++ level) violate the "as-if" rule; that is all transformations should produce a semantically equivalent code.

However, beyond the C++ level, when the code is optimized, further opportunities may arise.

As such, it really depends on the objects themselves: if the move-constructors/destructors have side effects, and (de)allocating memory is a side effect, then the optimization cannot occur. If you use only PODs, with default move-constructors/destructors, then it will probably be automatically optimized.