7.1.6.2 [dcl.type.simple] For an expression e, the type denoted by decltype(e) is defined as follows: — if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the … Read more
In this example, they mean the exact same thing. However, there are a few advantages to using the trailing return type form consistently (Phil Nash calls these “East End Functions”, since the return type is on the east end). Using parameters. Obviously when using parameters to determine the return type, you must use a trailing … Read more
decltype also considers whether the expression is rvalue or lvalue . Wikipedia says, The type denoted by decltype can be different from the type deduced by auto. #include <vector> int main() { const std::vector<int> v(1); auto a = v[0]; // a has type int decltype(v[0]) b = 1; // b has type const int&, the … Read more
What is type of auto in here ? The type is decltype(m_some_class) – I.e., the return value is of the same type as the variable m_some_class. Note that the function will return a copy of *this. If a reference to *this is wanted instead, you can use auto& or, since C++14, the more generic decltype(auto).
I think this runs into §7.1.6.4 [dcl.spec.auto]/p11: If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. You need the type of terminal to determine the type of the id-expression terminal in return terminal(func(term)); (edited, hat tip @Richard Smith), but at … Read more
In a nutshell, auto cannot be used in an effort to omit the actual types of function arguments, so stick with function templates and/or overloads. auto is legally used to automatically deduce the types of variables: auto i=5; Be very careful to understand the difference between the following, however: auto x=… auto &x=… const auto … Read more
The code is right. auto&& p = expr means the type of p is T&& where T will be inferred from expr. The && here indicates a rvalue reference, so e.g. auto&& p = 1; will infer T == int and thus the type of p is int&&. However, references can be collapsed according to … Read more
In C# var keyword works only locally inside function: var i = 10; // implicitly typed In C++ auto keyword can deduce type not only in variables, but also in functions and templates: auto i = 10; auto foo() { //deduced to be int return 5; } template<typename T, typename U> auto add(T t, U … Read more
Author answer: I just emailed Mr Van der Linden, and here is what he said: Yes, I agree with the people who answered on stack overflow. I don’t know for certain, because I never used the language B, but it seems highly plausible to me that “auto” ended up in C because it was in … Read more
auto is not a type so I’m not surprised this doesn’t work. But can’t you use decltype? [](auto&& v) { f(std::forward<decltype(v)>(v)); }(8); Scott Meyers has more details.