#include <iostream>
#include <string>
using namespace std;
template <typename T1>
class A
{
public://构造函数模板template <typename T2>A(T2 v1, T2 v2);//成员函数模板template <typename T3>void myft(T3 tmpt) {cout << tmpt << endl;}T1 m_ic;static constexpr int m_stcvalue = 200;
public:A(double v1, double v2){cout << "A(double v1, double v2)" << endl;}A(T1 v1, T1 v2){cout << "A(T1 v1, T1 v2)" << endl;}
public://拷贝构造函数模板template <typename U>A(const A<U>& other)//A(A<U>& other){cout << "A::A(const A<U>& other)" << endl;}//拷贝赋值运算符模板template <typename U>A<T1>& operator=(const A<U>& other)//A<T1>& operator=(A<U>& other){cout << "operator=(const A<U>& other)" << endl;return *this;}
};
//类外定义
template <typename T1>
template <typename T2>
A<T1>::A(T2 v1, T2 v2)
{cout << "A::A(T2,T2)执行了!" << endl;
}
int main()
{A<float> a(11.1f, 12.2f);a.m_ic = 16.2f;A<float> a1(a);return 0;
}

//拷贝构造函数模板与拷贝赋值运算符模板的参数如果不是const,则不会调用。

如果带上const,并且又要进行相应的调用，那么这里的a1的类型需要与a的类型不一致，比如是int.