《Effective Modern C++》中“Item 3: Understand decltype”章节读书笔记

《 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);
}