基于按值返回函数正确初始化对象

Question

我基本上有以下两个类，我使用return-by-value函数来创建对象。 在下面的Bar类中，我有两个Foo类成员对象。我怎样才能正确初始化每一个 两个对象分开？ 下面，我将对显示的编译器错误进行说明。

template < typename T >
class Foo{

    public:
        Foo( );
        Foo( const Foo<T> & );
        ~Foo();

        friend Foo<T> createFoo1( double, bool );
        friend Foo<T> createFoo2( double, bool );
        friend Foo<T> createFoo3( int );

    private:

        std::vector<T> m_data;

};

template < typename T >
Foo<T> createFoo1( double param1, bool param2 ){
    Foo<T> myFoo;

    // fill myFoo.
    return (myFoo);
}

template < typename T >
class Bar{

    public:

        Bar( );
        Bar( const Foo<T> &, const Foo<T> & );
        Bar( const Bar<T> & );
        ~Bar( );

        friend Bar<T> createBar1( double, bool );

    private:
        Foo<T> m_fooY;
        Foo<T> m_fooX;
};

template < typename T >
Bar<T> createBar1( double param1, bool param2 ){
    Bar<T> myBar( createFoo1<T>(param1, param2), createFoo1<T>(param1, param2) ); //OK
    return (myBar);

    //Bar<T> myBar;
    //myBar.m_fooY(createFoo1<T>(param1, param2)); // <- error C2064: term does not evaluate to a function taking 1 arguments
    //myBar.m_fooX(createFoo1<T>(param1, param2)); // <- error C2064: term does not evaluate to a function taking 1 arguments
    //return (myBar);
}

Answer 1

除了通过构造函数之外，您还可以设置字段m_fooX和m_fooY：

template < typename T >
Bar<T> createBar1( double param1, bool param2 ){
  Bar<T> myBar;
  myBar.m_fooY = createFoo1<T>(param1, param2);
  myBar.m_fooX = createFoo1<T>(param1, param2);
  return myBar;
}