decltype vs auto

Question

As I understand it, both decltype and auto will attempt to figure out what the type of something is.

If we define:

int foo () {
    return 34;
}

Then both declarations are legal:

auto x = foo();
cout << x << endl;

decltype(foo()) y = 13;
cout << y << endl;

Could you please tell me what the main difference between decltype and auto is?

Answer 1

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&`
    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&`
    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.