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 () {         


        

Answer 1

modify @Mankarse's example code,I think a better one blew:

#include <iostream>
int global = 0;
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 << "global: " << global << '\n'; //prints "global: 0"

    std::cout << "---\n";

    //a is an `int&`
    a = 10;

    std::cout << "a: " << a << '\n'; //prints "a: 10"
    std::cout << "b: " << b << '\n'; //prints "b: 2"
    std::cout << "global: " << global << '\n'; //prints "global: 10"

    return 0;

}

Answer 2

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.

Answer 3

Generally, if you need a type for a variable you are going to initialize, use auto. decltype is better used when you need the type for something that is not a variable, like a return type.

Answer 4

auto (in the context where it infers a type) is limited to defining the type of a variable for which there is an initializer. decltype is a broader construct that, at the cost of extra information, will infer the type of an expression.

In the cases where auto can be used, it is more concise than decltype, as you don't need to provide the expression from which the type will be inferred.

auto x = foo();                           // more concise than `decltype(foo()) x`
std::vector<decltype(foo())> v{ foo() };  // cannot use `auto`

The keyword auto is also used in a completely unrelated context, when using trailing return types for functions:

auto foo() -> int;

There auto is only a leader so that the compiler knows that this is a declaration with a trailing return type. While the example above can be trivially converted to old style, in generic programming it is useful:

template <typename T, typename U>
auto sum( T t, U u ) -> decltype(t+u)

Note that in this case, auto cannot be used to define the return type.