c++decltype关键字的用法

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&)

struct Point {

int x, y; // decltype(Point::x) is int, decltype(Point::y) is int

};

Widget w; // decltype(w) is Widget

if (f(w)) ... // decltype(f(w)) is bool

templateclass vector {

public:

...

T& operator[](std::size_t index);...

};

vector v; // decltype(v) is vector

if (v[0] == 0) ... // decltype(v[0]) is int&

template
auto authAndAccess(Container& c, Index i) -> decltype(c[i]) {

 authenticateUser();

 return c[i];
}

template
auto authAndAccess(Container& c, Index i) {

 authenticateUser();

 return c[i]; // return type deduced from c[i]
}

std::deque d;

...

authAndAccess(d, 5) = 10; //return d[5], then assign 10 to it; this won't compile!

template
decltype(auto) authAndAccess(Container& c, Index i){

 authenticateUser();

 return c[i];
}

Widget w;
const Widget& cw = w;
auto myWidget1 = cw; // auto type deduction: myWidget1's type is Widget

decltype(auto) myWidget2 = cw; // decltype type deduction: myWidget2's type is const Widget&

template
decltype(auto) authAndAccess(Container& c, Index i);

template
decltype(auto) authAndAccess(Container&& c,Index i);

template
decltype(auto)authAndAccess(Container&& c, Index i){

 authenticateUser();

 return std::forward(c)[i];
} // c++ 14版本

template
auto authAndAccess(Container&& c, Index i)
-> decltype(std::forward(c)[i])
{
 authenticateUser();

 return std::forward(c)[i];
} // c++ 11版本

decltype(auto) f1() { // decltype(x) is int, so f1 returns int
 int x = 0;
 ...
 return x;
}

decltype(auto) f2() { // decltype((x)) is int&, so f2 returns int&
 int x = 0;
 ...
 return (x);
}