1. variable
2. function
3. use typename will consider as a type
4. if it's a template, use template<T> name
C++11:
auto as a return value of function does not behave the same as auto variable declaration.
auto HA() ->decltype(T()) {}
auto must be plain type 'auto', can't be used as auto&& deduced type:
const int a; auto& tmp_1 = a; // reference to const int auto tmp_2 a; //auto is type int
difference between name and expression:
struct Test { int tmp; }; const Test t; decltype(t.tmp) tmp; // tmp is type int decltype((t.tmp)) tmp; // tmp is type const int&
also:
Why does C++ require a user-provided default constructor to default-construct a const object?
POD:
POD Type What are Aggregates and PODs and how/why are they special?
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