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
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 … Read more
auto was a keyword that C++ “inherited” from C that had been there nearly forever, but virtually never used because there were only two possible conditions: either it wasn’t allowed, or else it was assumed by default. The use of auto to mean a deduced type was new with C++11. At the same time, auto … Read more
By using auto&& var = <initializer> you are saying: I will accept any initializer regardless of whether it is an lvalue or rvalue expression and I will preserve its constness. This is typically used for forwarding (usually with T&&). The reason this works is because a forwarding reference, auto&& or T&&, will bind to anything. … Read more
auto can aid performance by avoiding silent implicit conversions. An example I find compelling is the following. std::map<Key, Val> m; // … for (std::pair<Key, Val> const& item : m) { // do stuff } See the bug? Here we are, thinking we’re elegantly taking every item in the map by const reference and using the … 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.
auto keyword does not work as a type for function arguments, in C++11. If you don’t want to use the actual type in lambda functions, then you could use the code below. for_each(begin(v), end(v), [](decltype(*begin(v)) it ){ foo( it + 5); }); The code in the question works just fine in C++ 14.
A range based loop could be a cleaner solution: for (const auto& i : a) { } Here, i is a const reference to an element of container a. Otherwise, if you need the index, or if you don’t want to loop over the entire range, you can get the type with decltype(a.size()). for (decltype(a.size()) … Read more
The difference is that in the first case auto is deduced to int* while in the second case auto is deduced to int, which results in both p1 and p2 being of type int*. The type deduction mechanism for auto is equivalent to that of template arguments. The type deduction in your example is therefore … Read more