#include <vector> 
#include <map> 
 
using namespace std; 
 
int main(int argc, char *argv[], char *env[]) 
{ 
// auto a;  // 错误，没有初始化表达式，无法推断出a的类型 
// auto int a = 10; // 错误，auto临时变量的语义在C++11中已不存在, 这是旧标准的用法。 
 
 // 1. 自动帮助推导类型 
 auto a = 10; 
 auto c = 'A'; 
 auto s("hello"); 
 
 // 2. 类型冗长 
 map<int, map<int,int> > map_; 
 map<int, map<int,int>>::const_iterator itr1 = map_.begin(); 
 const auto itr2 = map_.begin(); 
 auto ptr = []() 
 { 
 std::cout << "hello world" << std::endl; 
 }; 
 
 return 0; 
}; 
 
// 3. 使用模板技术时，如果某个变量的类型依赖于模板参数， 
// 不使用auto将很难确定变量的类型（使用auto后，将由编译器自动进行确定）。 
template <class T, class U> 
void Multiply(T t, U u) 
{ 
 auto v = t * u; 
} 

template <typename T1, typename T2> 
auto compose(T1 t1, T2 t2) -> decltype(t1 + t2) 
{ 
 return t1+t2; 
} 
auto v = compose(2, 3.14); // v's type is double 

auto k = 5; 
auto* pK = new auto(k); 
auto** ppK = new auto(&k); 
const auto n = 6; 

auto m; // m should be intialized 

auto int p; // 这是旧auto的做法。 

void MyFunction(auto parameter){} // no auto as method argument 
 
template<auto T> // utter nonsense - not allowed 
void Fun(T t){} 

int* p = new auto(0); //fine 
int* pp = new auto(); // should be initialized 
 
auto x = new auto(); // Hmmm ... no intializer 
 
auto* y = new auto(9); // Fine. Here y is a int* 
auto z = new auto(9); //Fine. Here z is a int* (It is not just an int) 

int value = 123; 
auto x2 = (auto)value; // no casting using auto 
 
auto x3 = static_cast<auto>(value); // same as above 

auto x1 = 5, x2 = 5.0, x3='r'; // This is too much....we cannot combine like this 

const int i = 99; 
auto j = i; // j is int, rather than const int 
j = 100 // Fine. As j is not constant 
 
// Now let us try to have reference 
auto& k = i; // Now k is const int& 
k = 100; // Error. k is constant 
 
// Similarly with volatile qualifer 

int a[9]; 
auto j = a; 
cout<<typeid(j).name()<<endl; // This will print int* 
 
auto& k = a; 
cout<<typeid(k).name()<<endl; // This will print int [9]