浅析C++模板类型中的原样转发和可变参数的实现

template <typename F, typename T1, typename T2>
void flip1(F f, T1 t1, T2 t2)
{
    f(t2, t1);
}

void ftemp(int v1, int &v2)
{
    cout << v1 << " " << ++v2 << endl;
}
void use_ftemp(){
    int j = 100;
    int i = 99;
    flip1(ftemp, j, 42);
    cout << "i is " << i << " j is " << j << endl;
}

template <typename F, typename T1, typename T2>
void flip2(F f, T1 &&t1, T2 &&t2)
{
    f(t2, t1);
}

int j = 100;
int i = 99;
flip2(ftemp, j, 42);
cout << "i is " << i << " j is " << j << endl;

void gtemp(int &&i, int &j)
{
    cout << "i is " << i << " j is " << j << endl;
}

int j = 100;
int i = 99;
// flip2(gtemp, j, 42) 会报错
// 因为42作为右值纯递给flip2，t2会被折叠为int&类型
// t2传递给gtemp第一个参数时，int&&无法绑定int&类型
//flip2(gtemp, i, 42);
cout << "i is " << i << " j is " << j << endl;

template <typename F, typename T1, typename T2>
void flip(F f, T1 &&t1, T2 &&t2)
{
    f(std::forward<T2>(t2), std::forward<T1>(t1));
}

void use_ftemp()
{
    int j = 100;
    int i = 99;
    flip(gtemp, i, 42);
    cout << "i is " << i << " j is " << j << endl;
}

//可变参数的函数模板
template <typename T>
ostream &print(ostream &os, const T &t)
{
    return os << t; //输出最后一个元素
}
template <typename T, typename... Args>
ostream &print(ostream &os, const T &t, const Args &...rest)
{
    os << t << ", ";
    return print(os, rest...);
}

void use_printtemp()
{
    int i = 100;
    string s = "hello zack!!!";
    print(cout, i, s, 42);
}