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.