《 Effective Modern C++》真是本好书，，讲的深入浅出，非常清晰

decltype

Given a name or an expression, decltype tells you the name’s or the expression’s type

const int i = 0;            // decltype(i) is const int  bool f(const Widget& w);    // decltype(w) is const Widget&                             // decltype(f) is bool(const Widget&)  template<typename T> class vector { public:     T& operator[](std::size_t index); }; vector<int> v;              // decltype(v) is vector<int>                             // decltype(v[0]) is int& 

如何查看 decltype(f)、decltype(v[0])的值？

template<typename T> class TD;                   // Type Displayer  TD<decltype(v[0])> x; 

编译如上代码，编译器会报错如下：

error: aggregate 'TD<int&> x' has incomplete type and cannot be defined 

由此可知 decltype(v[0])的结果是int&

decltype(auto)

auto specifies that the type is to be deduced, and decltype says that decltype rules should be used during the deduction

template<typename Container, typename Index> decltype(auto) authAndAccess(Container& c, Index i) {     authenticateUser();     return c[i]; }  

这里之所以使用 decltype(auto)而不是 auto，是因为绝大多数 containers-of-T 的[]运算符返回 T&，用 auto 的话会丢失&属性

decltype((name))

if an lvalue expression other than a name has type T, decltype reports that type as T&

int x = 3;      // decltype(x)   is int                 // decltype((x)) is int& 

注意：decltype(-x)等不会出现这种问题，因为-x等表达式都是右值，只有(x)既是表达式还是左值

这将导致下面两个函数的返回值类型不同

decltype(auto) f1() {     int x = 0;     return x; }  decltype(auto) f2() {     int x = 0;     return (x); }