Using auto in a lambda function

This was discussed briefly by Herb Sutter during an interview. Your demand for auto arguments is in fact no different from demanding that any function should be declarable with auto, like this:

auto add(auto a, auto b) -> decltype(a + b) { return a + b; }

However, note that this isn't really a function at all, but rather it's a template function, akin to:

template <typename S, typename T>
auto add(S a, T b) -> decltype(a + b) { return a + b; }

So you are essentially asking for a facility to turn any function into a template by changing its arguments. Since templates are a very different sort of entity in the type system of C++ (think of all the special rules for templates, like two-phase look-up and deduction), this would be radical design change with unforeseeable ramifications, which certainly isn't going to be in the standard any time soon.

C++14 allows lambda function (Generic lambda function) parameters to be declared with the auto.

auto multiply = [](auto a, auto b) {return a*b;};

For details: http://en.cppreference.com/w/cpp/language/lambda

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.