decltype typename

decltype((variable))总是引用类型，但是decltype(variable)只有当variable是引用类型时才是引用类型。

#include <iostream>
#include <typeinfo>

using std::cin;
using std::cout;
using std::endl;

auto f(auto x)
{ return x+1;}
auto f(auto x,auto y)->decltype(y){
    return x-y;
}
 
double func()
{
    cout << "func executed." << endl;
    return 2.5+3.6;
}

int main()
{
    decltype(func()) sum;
    cout << typeid(sum).name() << endl;
    
    const int ci = 0, &cj = ci;
    decltype(ci) x = 0;
    decltype(cj) y = x;
    //decltype(cj) z; // compile error: ‘z’ declared as reference but not initialized
    cout << typeid(x).name() << endl;
    cout << typeid(y).name() << endl;
    
    int i = 10, *p = &i, &r = i;
    decltype(r + 0) b;
    //decltype(*p) c; // compile error: ‘c’ declared as reference but not initialized
    cout << typeid(b).name() << endl;
    
    decltype(i) u;
    //decltype((i)) v; // compile error: ‘v’ declared as reference but not initialized
    
    return 0;
}

 

转载于:https://www.cnblogs.com/huashiyiqike/p/3927894.html