auto-typed variables are deduced by the compiler according to the type of their initializer.

auto a = 3.14; // double
auto b = 1; // int
auto& c = b; // int&
auto d = { 0 }; // std::initializer_list<int>
auto&& e = 1; // int&&
auto&& f = b; // int&
auto g = new auto(123); // int*
const auto h = 1; // const int
auto i = 1, j = 2, k = 3; // int, int, int
auto l = 1, m = true, n = 1.61; // error -- `l` deduced to be int, `m` is bool
auto o; // error -- `o` requires initializer

Extremely useful for readability, especially for complicated types:

std::vector<int> v = ...;
std::vector<int>::const_iterator cit = v.cbegin();
// vs.
auto cit = v.cbegin();

Functions can also deduce the return type using auto. In C++11, a return type must be specified either explicitly, or using decltype like so:

template <typename X, typename Y>
auto add(X x, Y y) -> decltype(x + y) {
  return x + y;
}
add(1, 2); // == 3
add(1, 2.0); // == 3.0
add(1.5, 1.5); // == 3.0

The trailing return type in the above example is the declared type (see section on decltype) of the expression x + y. For example, if x is an integer and y is a double, decltype(x + y) is a double. Therefore, the above function will deduce the type depending on what type the expression x + y yields. Notice that the trailing return type has access to its parameters, and this when appropriate.