views:

196

answers:

2

I am really confused now on how and which method to use to return object from a function. I want some feedback on the solutions for the given requirements.

Scenario A: The returned object is to be stored in a variable which need not be modified during its lifetime. Thus,

const Foo SomeClass::GetFoo() {
 return Foo(); 
}

invoked as:

someMethod() {
 const Foo& l_Foo = someClassPInstance->GetFoo();
//...
}

Scneraio B: The returned object is to be stored in a variable which will be modified during its lifetime. Thus,

void SomeClass::GetFoo(Foo& a_Foo_ref) {
     a_Foo_ref = Foo(); 
    }

invoked as:

someMethod() {
 Foo l_Foo;
 someClassPInstance->GetFoo(l_Foo);
//...
}

I have one question here: Lets say that Foo cannot have a default constructor. Then how would you deal with that in this situation, since we cant write this anymore:

Foo l_Foo

Scenario C:

Foo SomeClass::GetFoo() {
 return Foo(); 
}

invoked as:

someMethod() {
 Foo l_Foo = someClassPInstance->GetFoo();
//...
}

I think this is not the recommended approach since it would incur constructing extra temporaries.

What do you think ? Also, do you recommend a better way to handle this instead ?

+4  A: 

Out of the three scenarios, number 3 is the ideomatic method, and the one you should probably use. You won't be paying for extra copies because the compiler is free to use copy elision to avoid the copy if possible.

Secnario A is wrong. You end up with a reference to a temporary which is destroyed when the statement containing the function call finishes. Okay, Scenario A is not wrong, but you should still use Scenario C.

Secnario B works just fine, but:

Lets say that Foo cannot have a default constructor. Then how would you deal with that in this situation, since we cant write this anymore: Foo l_Foo.

Foo has to have a default constructor. Even if you don't give it one, the compiler must for you. Assuming you declare a private constructor, there's no way you can use this calling method. You need to either make Foo default constructable or use Secnario C.

Billy ONeal
`const Foo SomeClass::GetFoo()` :: Scenario A :: I am not returning the reference to a temporary here..
brainydexter
@brainydexter: Oops.. you are correct. However, I believe it's still wrong. I'm not 100% sure, but I believe the temporary will be destroyed despite having a reference to it. You need to have the actual object around.
Billy ONeal
scenario A is fine but really no improvement over C. There is a special rule in C++ that extends the life-time of the temporary object in this case.
sellibitze
@Billy: You asked for it :-) I asked this earlier and it is valid.You can read about it here:http://stackoverflow.com/questions/2615162/return-value-not-a-reference-from-the-function-bound-to-a-const-reference-in-thttp://herbsutter.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/
brainydexter
@brainydexter: D'oh! Fixed.
Billy ONeal
+6  A: 

First, let's look into the things that come into play here:

(a) Extending lifetime of a temporary when it's used to initialize a reference - I learnt about it in this publication by Andrei Anexandrescu. Again, it feels weird but useful:

class Foo { ... }

Foo GetFoo() { return Foo(); }  // returning temporary

void UseGetFoo()
{
   Foo const & f = GetFoo();
   // ... rock'n'roll ...
   foo.StillHere();
}

The rule says that when a reference is initialized with a temporary, the temporary's lifetime is extended until the reference goes out of scope. (this reply quotes the canon)

(b) Return Value Optimization - (wikipedia) - the two copies local --> return value --> local may be omitted under circumstances. That's a surprising rule, as it allows the compiler to change the observable behavior, but useful.

There you have it. C++ - weird but useful.


So looking at your scenarios

Scenario A: you are returning a temporary, and bind it to a reference - the temporary's lifetime is extended to the lifetime of l_Foo.

Note that this wouldn't work if GetFoo would return a reference rather than a temporary.

Scenario B: Works, except that it forces a Construct-Construct-Copy-Cycle (which may be much more expensive than single construct), and the problem you mention about requiring a default constructor.

I wouldn't use that pattern to create a object - only to mutate an existing one.

Scenario C: The copies of temporaries can be omitted by the compiler (as of RVO rule). There is unfortunately no guarantee - but modern compilers do implement RVO.

Rvalue references in C++ 0x allows Foo to implement a resource pilfering constructor that not only guarantees supression of the copies, but comes in handy in other scenarios as well.

(I doubt that there's a compiler that implements rvalue references but not RVO. However there are scenarios where RVO can't kick in.)


A question like this requires mentioning smart pointers, such as shared_ptr and unique_ptr (the latter being a "safe" auto_ptr). They are also in C++ 0x. They provide an alternate pattern for functions creating objects.


peterchen
brainydexter
Also, look at my last comment for original question.
brainydexter
@brainydexter: Yes, that's a typical use scenario for a reference. It requires some documentation (is it out or in/out etc.). Thanks for the reference, I'll update with a link. --- I've recently toyed around with rvalue references - it was a good exercise for me, too.
peterchen