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
auto newvar1 = myvector; // vs: auto *newvar2 = myvector; Both of these are the same and will declare a pointer to std::vector<MyClass> (pointing to random location, since myvector is uninitialized in your example and likely contains garbage). So basically you can use any one of them. I would prefer auto var = getVector(), but … Read more
You cannot use auto to store parser expressions¹ Either you need to evaluate from the temporary expression directly, or you need to assign to a rule/grammar: const qi::rule<std::string::const_iterator, qi::space_type> doubles_parser_local = qi::double_ >> *(‘,’ >> qi::double_); You can have your cake and eat it too on most recent BOost versions (possibly the dev branch) there … Read more
There’s surprisingly little in the standard about this. About all we hear about redeclaration is: [C++11: 3.1/1]: A declaration (Clause 7) may introduce one or more names into a translation unit or redeclare names introduced by previous declarations. [..] and the only relevant part of auto‘s semantics: [C++11: 7.1.6.4/3]: Otherwise, the type of the variable … Read more
auto and auto && cover most of the cases: Use auto when you need a local copy. This will never produce a reference. The copy (or move) constructor must exist, but it might not get called, due to the copy elision optimization. Use auto && when you don’t care if the object is local or … Read more
In C++11, there are two syntaxes for function declaration: return-type identifier ( argument-declarations… ) and auto identifier ( argument-declarations… ) -> return_type They are equivalent. Now when they are equivalent, why do you ever want to use the latter? Well, C++11 introduced this cool decltype thing that lets you describe type of an expression. So … Read more
Because std::vector<bool> is not a container ! std::vector<T>‘s iterators usually dereference to a T&, which you can bind to your own auto&. std::vector<bool>, however, packs its bools together inside integers, so you need a proxy to do the bit-masking when accessing them. Thus, its iterators return a Proxy. And since the returned Proxy is an … Read more
Read this article: Appearing and Disappearing consts in C++ Type deduction for auto variables in C++0x is essentially the same as for template parameters. (As far as I know, the only difference between the two is that the type of auto variables may be deduced from initializer lists, while the types of template parameters may … 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
Yes, this is an extension. It’s likely to be added to C++17 as part of the ‘concepts’ proposal, I believe.