c++ - C++显式构造函数不阻止将double转换为int

Question

我有一个从int到C的构造函数，从double到一个。

我让第一个执行隐式类型转换，但是使用关键字explicit阻止第二个。

但是，不幸的是，出现了从double到int的隐式转换。
我可以以某种方式阻止它吗？

这是一个简化的例子

//g++  5.4.0
#include <iostream>
using namespace std;

class C{
    int* tab;
    public:
    C():tab(nullptr){ cout<<"(void)create zilch\n"; }
    C(int size):tab(new int[size]){ cout<<"(int)create " << size << "\n"; }
    explicit C(double size):tab(new int[(int)size]){ cout<<"(double)create " << size << "\n"; }
    ~C(){ if(tab) {cout<<"destroy\n"; delete[] tab;} else cout <<"destroy zilch\n"; }
};    

int main()
{
    cout << "start\n";
    {
        C o1(1);
        C o2 = 2; //ok, implicit conversion allowed
        C o3(3.0);
        C o4 = 4.0; //ko, implicit conversion to double blocked... but goes to int 
    }
    cout << "stop\n";
}

//trace 
//
//start
//(int)create 1
//(int)create 2
//(double)create 3
//(int)create 4
//destroy
//destroy
//destroy
//destroy
//stop

Answer 1

好! Eljay 在注释中获得了更快的速度，但这是最终代码，将使您在尝试隐式使用double时出现编译错误

#include <iostream>
using namespace std;

class C{
    int* tab;

public:
    //THE TRICK: block any implicit conversion by default
    template <class T> C(T) = delete;

    C():tab(nullptr){ cout<<"(void)create zilch\n"; }
    C(int size):tab(new int[size]){ cout<<"(int)create " << size << "\n"; }
    explicit C(double size):tab(new int[(int)size]){ cout<<"(double)create " << size << "\n"; }
    ~C(){ if(tab) {cout<<"destroy\n"; delete[] tab;} else cout <<"destroy zilch\n"; }
};

int main()
{
    cout << "start\n";
    {
        C o1(1);
        C o2 = 2; //ok, implicit conversion allowed
        C o3(3.0);
        C o4 = 4.0; //ko, implicit conversion to other types deleted
        C o5 = (C)5.0; //ok. explicit conversion
    }
    cout << "stop\n";
}