C++ : copy-and-swap idiom, 替代构造函数

Question

注意:此问题遵循 a previous one , 我希望仍然可以将其作为一个新问题提出。

我正在尝试为树类实现“三个半大规则”( copy-and-swap 习语)，它看起来像这样:

class Tree
{
    friend void swap(Tree &first, Tree &second); // Swap function

public:
    Tree(const double &a, const double &b, int depth); // Public constructor (derived from the default (private) constructor)
    Tree(const Tree &other); // Copy constructor
    ~Tree(); // Destructor
    Tree & operator=(Tree other); // Copy-assignement operator


private:        
    Tree(double *a, double *b, int depth, int maxDepth); // Default (private) constructor

    double *a, *b;
    int depth, maxDepth;    
    Tree *leftChild, *rightChild;
};

我一直在努力关注this guideline .这是我的复制赋值运算符的样子:

Tree & Tree::operator=(Tree other)
{
    swap(*this, other);
    return *this;
}

我很难让我的公共(public)构造函数工作。有人建议我做类似的事情:

Tree::Tree(const double &a, const double &b, int depth)
{
    double aTemp(a), bTemp(b);
    swap(*this, Tree(&aTemp, &bTemp, depth, depth));
}

我不确定这个想法是否可行。在任何情况下，我都会从编译器中得到以下错误:

invalid initialization of non-const reference of type 'Tree&' from an rvalue of type 'Tree'
in passing argument 2 of 'void swap(Tree&, Tree&)'

我尝试了以下想法，我认为它可行:

Tree::Tree(const double &a, const double &b, int depth)
{
    double aTemp(a), bTemp(b);
    *this = Tree(&aTemp, &bTemp, depth, depth);
}

但是好像也没有用。我认为问题在于，当我调用复制赋值运算符 (*this = Tree(&aTemp, &bTemp, depth, depth)) 时，应该调用复制构造函数(因为复制的参数-assignement operator is passed by value)，但似乎这没有发生。我不明白为什么。

在此先感谢您的帮助!

Answer 1

invalid initialization of non-const reference of type 'Tree&' from an rvalue of type 'Tree' in passing argument 2 of 'void swap(Tree&, Tree&)'

C++ 不允许通过非const 引用传递匿名对象。目的是防止调用者意外丢弃写入引用参数的函数的结果。

你可以这样做:

Tree::Tree(const double &a, const double &b, int depth)
{
    double aTemp(a), bTemp(b);
    Tree temp(&aTemp, &bTemp, depth, depth);
    swap(*this, temp);
}

But it does not seem to be working either. I think the problem is that when I call the copy-assignment operator (*this = Tree(&aTemp, &bTemp, depth, depth)), the copy constructor should be called (since the argument of the copy-assignement operator is passed by value), but it seems that this is not happening. I do not understand why.

您如何确定它不起作用？编译器可能会删除 拷贝以避免做不必要的工作。 ( That is why your copy-assignment operator takes the argument by value. )

顺便说一句，如果你的编译器支持 C++11，你可以使用 delegating constructors相反。