c++ - 为什么构造函数优先于用户定义的运算符？

Question

我正在尝试理解 static_cast。我有两个类(class) From 和 To。我正在尝试将 From 转换为 To。我有一个 To 的构造函数，它采用 From。我还有一个用户定义的转换运算符，用于将 From 对象转换为 To。为什么构造函数优先于赋值运算符？

class From;
class To
{
public:
    int y;
    To() { cout << "In To default constructor" << endl; }
    To(const To& sl2) { cout << "In To copy constructor" << endl; }
    void operator=(const To& sl2) { cout << "In To assignment operator" << endl; }
    ~To() { cout << "In To destructor" << endl; }
    To(const From& sl1) { cout << "In From-To constructor" << endl;}
};

class From
{
public:
    int x;
    From() { cout << "In From default constructor" << endl; }
    From(const From& sl2) { cout << "In From copy constructor" << endl; }
    void operator=(const From& sl2) { cout << "In From assignment operator" << endl; }
    ~From() { cout << "In From destructor" << endl; }
    operator To() const { cout << "Converting From to To" << endl; return To(); } 
};

int _tmain(int argc, _TCHAR* argv[])
{
    From from;
    To to;
    to = static_cast<To>(from);
}

Answer 1

“因为标准这么说”对你来说是一个有效的答案吗？

static_cast<T>(e)与 T t(e); 效果相同与 t是类型转换的结果。这遵循直接初始化的规则，它在考虑用户定义的转换序列之前通过重载决议考虑适用的构造函数。

如果您放心，我可以提供相关的标准报价。