This code uses 2 new features from C++11 standard the auto keyword, for type inference, and the range based for loop. Using just auto this can be written as (thanks Ben) for (auto it=mymap.begin(); it!=mymap.end(); ++it) Using just range for this can be written as for (std::pair<const char,int>& x: mymap) { std::cout << x.first << … Read more
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
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
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
The rule for prohibiting non-static members is in 7.1.6.4 clause 4: The auto type-specifier can also be used in declaring a variable in the condition of a selection statement (6.4) or an iteration statement (6.5), in the type-specifier-seq in the new-type-id or type-id of a new-expression (5.3.4), in a for-range-declaration, and in declaring a static … 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
To make long story short: a braced initializer expression {} has no type by itself auto has to infer type information int{3} obviously means “create an int var with value taken from initializer list”, thus its type is just int and can be used in any wider context (int i = int{3} will work and … Read more
The declarations are exactly equivalent. auto works (almost) the same as template type deduction. Putting the star explicitly makes the code a bit easier to read, and makes the programmer aware that bar2 is a pointer.
Sorry, but I just think the best suggestion is not using auto at all, since you want to perform a (implicitly valid) type conversion. auto is meant for deducing the exact type, which is not what you want here. Just write it this way: std::map<std::string, int>::const_iterator city_it = usa.find(“New York”); As correctly pointed out by … Read more