转换构造函数不接受支撑列表

Question

所以当我搞乱构造函数时，我制作了这段代码：

class super
{
public:
int duper, duperer;

friend ostream& operator << (ostream&, super&);
friend super operator+ (super& uno, super& dos);
friend int main();
super() {};
super(int);
super(int, int);
};

ostream& operator << (ostream& out, super& Sout)
{
    cout << "The " << Sout.duper << " and the " << Sout.duperer << " are super!" << endl;
    return out;
}

super operator+ (super& uno, super& dos)
{
    return { uno.duper + dos.duper , uno.duperer + dos.duperer };
}


super::super(int duper) : duper(duper) {};
super::super(int dupster, int pupster) : duper(pupster), duperer(dupster)     {};

int main()
{

super Supper = { -1, 0 }, Copper(1, 1);
Supper = super(3, 3) + Copper;
cout << Supper << "!!!\n";
}

一切正常，初始化列表每次都转换为我的类型的对象。但是，当我这样做时（更改主要支撑列表的构造函数）：

class super
{
public:
int duper, duperer;

friend ostream& operator << (ostream&, super&);
friend super operator+ (super& uno, super& dos);
friend int main();
super() {};
super(int);
super(int, int);
};

ostream& operator << (ostream& out, super& Sout)
{
    cout << "The " << Sout.duper << " and the " << Sout.duperer << " are super!" << endl;
    return out;
}

super operator+ (super& uno, super& dos)
{
    return { uno.duper + dos.duper , uno.duperer + dos.duperer };
}


super::super(int duper) : duper(duper) {};
super::super(int dupster, int pupster) : duper(pupster), duperer(dupster)     {};

int main()
{

super Supper = { -1, 0 }, Copper(1, 1);
Supper = {3, 3} + Copper;
cout << Supper << "!!!\n";
}

......代码没有编译。这次为什么没有支撑列表转换为对象？

非常感谢，请不要对我在这里使用的名字感到生气。我只是想确保一切都让人感到困惑。