move constructor not being called as expected

Question

I'm new to C++0x and I'm trying to wrap my head around rvalue references and move constructors. I'm using g++ 4.4.6 with -std=c++0x, and I'm confused by the following piece of code:

    class Foo 
    {
    public:
      Foo() 
        : p( new int(0) )
      {
        printf("default ctor\n");
      }

      Foo( int i )
        : p( new int(i) )
      {
        printf("int ctor\n");
      }

      ~Foo() 
      {
        delete p;
        printf("destructor\n");
      }

      Foo( const Foo& other ) 
        : p( new int( other.value() ) )
      {
        printf("copy ctor\n");
      }


      Foo( Foo&& other )
        : p( other.p )
      {
        printf("move ctor\n");
        other.p = NULL;
      }

      int value() const 
      {
        return *p;
      }

    private:
      // make sure these don't get called by mistake
      Foo& operator=( const Foo& );
      Foo& operator=( Foo&& );

      int* p;
    };


    Foo make_foo(int i) 
    {
      // create two local objects and conditionally return one or the other
      // to prevent RVO
      Foo tmp1(i);
      Foo tmp2(i);

      // With std::move, it does indeed use the move constructor
      //  return i ? std::move(tmp1) : std::move(tmp2);
      return i ? tmp1 : tmp2;

    }


    int main(void) 
    {
      Foo f = make_foo( 3 );

      printf("f.i is %d\n", f.value());

      return 0;
    }

I find that as written, the compiler uses the copy constructor build the object in main(). When I use the std::move line inside make_foo(), then the move constructor is used in main(). Why is std::move necessary inside make_foo()? I would think that although tmp1 and tmp2 are named objects inside make_foo(), when they're returned from a function they should become temporaries.

Answer 1

This is your problem:

return i ? tmp1 : tmp2;

A local variable in a function will only be moved from in the return statement if the return statement is just return var;. If you want to do that test you will need to use an if:

if (i) {
   return tmp1;
} else {
   return tmp2;
}

The citation is a bit convoluted, but it is in 12.8/31 and 12.8/32

12.8/32 When the criteria for elision of a copy operation are met or would be met save for the fact that the source object is a function parameter, and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue [...]

That is even if the expression is an lvalue, it will be considered to be an rvalue when the criteria in 12.8/31 is met, the second option in that block is:

12.8/31 in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object (other than a function or catch-clause parameter) with the same cv-unqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value.

Which determines that return tmp; allows for copy elision, but return (cond?tmp:tmp); doesn't.

Note that for the compiler to generate an implicit std::move in the return statement, the returned object must be a candidate for elision unless the it is also an argument to the function. Using the conditional operation inhibits copy elision, and at the same time inhibits the compiler from doing moving out of your objects. That second case might be simpler to code:

Foo make_foo(Foo src) {
   return src;           // Copy cannot be elided as 'src' is an argument
}