Thread: why visual studio does not optimize constructor in this case

Hello everyone,


Why visual studio does not optimize constructor in this case? I do not understand what the MSDN mentioned,

if use different named object, compiler can not optimize. Why?

http://msdn2.microsoft.com/en-us/lib...57(vs.80).aspx

Code:

#include <stdio.h>
class RVO
{
public:
       
            RVO(){printf("I am in constructor\n");}
            RVO (const RVO& c_RVO) {printf ("I am in copy constructor\n");}
            int mem_var;       
};
RVO MyMethod (int i)
{
            RVO rvo;
            rvo.mem_var = i;
      if (rvo.mem_var == 10)
         return (RVO());
            return (rvo); 
}
int main()
{
            RVO rvo;
            rvo=MyMethod(5);
}

Output is,

I am in constructor
I am in constructor
I am in copy constructor

My expected output is,

I am in constructor
I am in constructor


thanks in advance,
George

Thanks laserlight,


I do not know why diferent execution path will disable compiler from optimization. In the sample, there are two execution paths to return,

1. return (RVO());

In case (1), compiler could optimize by saving the creating of the temporary object and assign to the extern rvo object directly.

2. return (rvo);

In this case (2), compiler could optimize by not creating the temporary object for the return value, and assign the inner rvo to outter rvo directly.

Why compiler can not optimize it?


regards,
George

Originally Posted by laserlight

If we look at it from a language lawyer point of view, the C++ Standard states that "in such cases, the implementation treats the source and target of the omitted copy operation as simply two different ways of referring to the same object". So, with different paths returning different named objects, which object to be returned should the implementation treat as the object from the caller? Since this cannot be determined at compile time, the return value optimisation cannot take place.

Thanks Elysia,


Why compiler can not optimize code for all the return paths? As I mentioned, in either return path, we can optimize the code by saving the time to create temporary object.

Anything wrong in my analysis?

Originally Posted by Elysia

If the return paths behave differently, then the compiler cannot know what code to generate at compile time and thus cannot optimize it at all. The compiler doesn't know which return path will be taken and cannot guess, so it cannot generate code because most likely it will generate faulty code.

regards,
George

Thanks Elysia,


But I do not know why which named variable/return path matters the optimization -- the optimization only happens when the function returns and compiler only needs to insert optimization code at the place where there is a return statement -- if there are multiple returns, the compiler could put multiple optimization into multiple return statement.

I think no matter which one will return, either rvo or RVO(), the return value is a temporary obejct, and compiler could optimize it by using assignment operator on the outside rvo object instance directly without creating the temporary object, right?

Why the named variable/return path matters? Could you provide more information about your analysis please?

Originally Posted by Elysia

I don't believe that's the problem. The compiler has to make sure the generated code works, and it cannot analyze for all situations when there are multiple returns paths or the compiler doesn't support it. I'm thinking it's the code when assigning the object to the temporary variable that cannot be optimized due to the different return paths, and thus the compiler cannot optimize anything.

regards,
George

Hi iMalc,


Any more specific analysis for the issue why compiler does not optimize the code in this case? :-)

Originally Posted by iMalc

This is why at some point in every programmers career they should try writing a compiler. It help you appreciate how difficult these things are. It teaches you about what is easy or difficult to optimise.

regards,
George

Hi CornedBee,


Thanks for sharing your perspective. But I do not agree with you. :-)

I think either rvo or RVO() is returned, compiler could optimize by not generating code to put a temporary object on the return stack by creating the copy constructor of the temporary object, and optimize it by invoking the outter rvo (in main)'s assignment operator directly.

This is the point why I am confused. I am not sure why compiler can not optimize in a similar way as I mentioned above.

I have also performed some further testing, if we change return RVO() to another named rvo instance, but different return path using different named rvo instance, the code can not be optimized either.

Code:

class RVO 
{ 
public: 

RVO(){printf("I am in constructor\n");} 
RVO (const RVO& c_RVO) {printf ("I am in copy constructor\n");} 
int mem_var; 
}; 
RVO MyMethod (int i) 
{ 
RVO rvo1;
RVO rvo2;
rvo1.mem_var = i; 
rvo2.mem_var = i; 
if (rvo1.mem_var == 10) 
return (rvo1); 
// return (RVO()); 
return (rvo2); 
} 
int main() 
{ 
	RVO rvo;
	rvo = MyMethod(5);

	return 0;
}

Output is,

--------------------
I am in constructor
I am in constructor
I am in constructor
I am in copy constructor
--------------------

Originally Posted by CornedBee

RVO and NRVO work by constructing the object that is to be returned directly in the space provided by the caller, instead of constructing the object in the local variables of the function and then copying it over to the caller space.

Let's look at the sample.

Code:

RVO MyMethod (int i)
{
    RVO rvo;
    rvo.mem_var = i;
    if (rvo.mem_var == 10)
        return (RVO());
    return (rvo); 
}

The issue is this: the rvo object must be constructed before the assignment to mem_var. Where should the compiler construct the object? If NRVO takes effect, the object is constructed directly in the space provided by the caller.
However, what if the alternate path is taken? Then the object that actually should be constructed in this space is not rvo, but the unnamed temporary that should be returned. If NRVO had been applied, this space would already be taken by rvo, leading to a conflict.
That's why rvo has to be a separate variable in the function's own space. NRVO cannot be applied. RVO for the unnamed temporary can still be applied.

regards,
George