根据按值返回函数正确初始化对象

Correct initialization of objects based on return-by-value functions

本文关键字：初始化 对象 函数 返回      更新时间：2023-10-16

我基本上有以下两个类，我使用按值返回函数来创建对象。在下面的Bar类中，我有两个Foo类成员对象。我怎样才能正确地初始化它们呢两个物体分开?下面，我将给出显示的编译器错误的示例。

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);
}

下面是设置m_fooX和m_fooY字段的方法，而不是通过构造函数:

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;
}
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