decltype gives the declared type of the expression that is passed to it. auto does the same thing as template type deduction. So, for example, if you have a function that returns a reference, auto will still be a value (you need auto& to get a reference), but decltype will be exactly the type of the return value.
#include <iostream>
int global{};
int& foo()
{
return global;
}
int main()
{
decltype(foo()) a = foo(); //a is an `int&`
// decltype(auto) a = foo(); alternatively, since C++14
auto b = foo(); //b is an `int`
b = 2;
std::cout << "a: " << a << '\n'; //prints "a: 0"
std::cout << "b: " << b << '\n'; //prints "b: 2"
std::cout << "---\n";
decltype(foo()) c = foo(); //c is an `int&`
// decltype(auto) c = foo(); alternatively, since C++14
c = 10;
std::cout << "a: " << a << '\n'; //prints "a: 10"
std::cout << "b: " << b << '\n'; //prints "b: 2"
std::cout << "c: " << c << '\n'; //prints "c: 10"
}
Also see David Rodríguez’s answer about the places in which only one of auto or decltype are possible.