auto关键字&decltype关键字

#include <iostream>
 
using namespace std;

template<class T, class U>
auto add(T t, U u)  -> decltype(t + u) //C++11 不加 -> decltype(t + u)会编译失败  C++14 auto才支持函数推导
{ 
    return t + u; 
} // 返回类型是 operator+(T, U) 的类型

int test(int a, int b)
{   
    return a * b;
}

int main()
{
    auto a = 1 + 2;
    auto b = add(2, 2);
    auto c = add(2, 2.3);
    auto d = {1.0, 2.0};
    auto &e = a; //e为a的 引用
    auto f = &e; //f为指向e 也就是a的 指针
    auto g = "111", h = "222";
    auto i = &f;//i是指向f的指针

    auto (*p)(int, int) -> int; //声明p为一个 int (int, int)的函数
    p = test;
    //auto (*q)(int, int) -> auto = p;//声明 q 为指向【返回 T 的函数】的指针  其中 T 从 p 的类型推导 C++14 可用
    //cout<<"q = "<<q(6, 6)<<endl;
    cout<<"p = "<<p(3, 3)<<endl; //9

    //  auto x;     // 在 C 合法，在 C++ 错误

    e++;
    *f = 30;

    cout << "a = "<<a<<" addr = "<<&a<<" "<<typeid(a).name() << endl; //30
    cout << typeid(b).name() << endl;//int
    cout << typeid(c).name() << endl;//double
    cout << typeid(d).name() << endl; //std::initializer_list<double> 
    cout << "e = "<<e<<" addr = "<<&e<<" "<<typeid(e).name() << endl; //30
    cout << "f = "<<*f<<" addr = "<<f<<" "<<typeid(f).name() << endl; //30
    cout << "g = "<<g<<" "<<typeid(g).name() << endl;
    cout << "h = "<<h<<" "<<typeid(h).name() << endl;  //类型为PKc  pointer(P) const(K) char(c)
    cout << "i = "<<**i<<" addr = "<<*i<<" "<<typeid(i).name() << endl;// PPi
}

#include <iostream>

using namespace std;

struct A { double x; };
const A* a;

decltype(a->x) y;       // y 的类型是 double（其声明类型）（右值表达式） a->x代表x的值
decltype((a->x)) z = 6.66; // z 的类型是 const double&（左值表达式） (a->x)代表double这个类型 

template<typename T, typename U>
auto add(T t, U u) -> decltype(t + u) // 返回类型依赖于模板形参
{
    return t+u;
}

int main()
{
    decltype(3 + 3) a = 6;
    y = 5.55;
    cout << "y = "<<y<<" addr = "<<&y<<" "<<typeid(y).name() << endl;

    decltype(&y) p = new double;//p是 *double  但是为NULL 需要new
    *p = 7.77;
    cout << "p = "<<*p<<" addr = "<<p<<" "<<typeid(p).name() << endl;
    delete p;

    double* d = &y;
    //(p) 为左值表达式
    decltype((p)) q = d; //q 是 double *& 类型  指向double类型的指针的引用 不能直接赋值&y 因为&y为右值表达式
    cout << "q = "<<*q<<" addr = "<<q<<" "<<typeid(q).name() << endl;

    decltype(add(4, 5.0)) k = 8.88; //k的类型为double  add(4, 5.0)为右值表达式
    cout << "k = "<<k<<" addr = "<<&k<<" "<<typeid(k).name() << endl;

    decltype((add(3, 6))) s = 10; //s的类型为int (add(3, 6))也为右值表达式
    cout << "s = "<<s<<" addr = "<<&s<<" "<<typeid(s).name() << endl;

    cout << "a = "<<a<<" addr = "<<&a<<" "<<typeid(a).name() << endl;
    cout << "z = "<<z<<" addr = "<<&z<<" "<<typeid(z).name() << endl;
    cout << "y = "<<y<<" addr = "<<&y<<" "<<typeid(y).name() << endl;
}