Why “auto” is not acceptable as lambda parameter

问题

Why does this code make a compile error?

std::find_if(std::begin(some_list), std::end(some_list), [](const auto& item){
//some code
});

The error of course at "auto"? why is not possible to know the type automatically ? thanks

回答1:

This is because as of C++11, lambda functions in C++ cannot be defined generically, therefore you cannot declare a parameter using auto. This has been added in the C++14 (and is already supported by some compilers).

However, you can achieve the same thing in C++11 using decltype(), in you case:

std::find_if(std::begin(some_list), std::end(some_list), [](decltype(*some_list.begin())& item){
        return item > 4;

来源：https://stackoverflow.com/questions/32646362/why-auto-is-not-acceptable-as-lambda-parameter