Variadic lambda vs variadic template function：严格等效？

Question

Consider this code snippet

#include <chrono>
auto wrapAndCallLambda = 
[] (auto fn, auto &&... params)
{
    const auto start = std::chrono::system_clock::now();

    auto result = fn(std::forward<decltype(params)>(params)...);
    const auto end = std::chrono::system_clock::now();

    const auto elapsed = end - start;
    std::cout << "Elapsed time: " << elapsed.count() << "s";
    return result;
};


template <typename Fn, typename... Args>
auto wrapAndCall(Fn fn, Args &&... params)
 {
    const auto start = std::chrono::system_clock::now();

    auto result = fn(std::forward<decltype(params)>(params)...);
    const auto end = std::chrono::system_clock::now();

    const auto elapsed = end - start;
    std::cout << "Elapsed time: " << elapsed.count() << "s";
    return result;
 }

Is it stricly equivalent? Is there any consideration to have in mind (not related to opinion and style) guiding which one to choose? Code bloating? Performances? Compile time if used extensively?

Answer 1

wrapAndCallLambda is an actual object, whereas wrapAndCall is a template.

This means that you can pass wrapAndCallLambda directly to other functions, and you have to either explicitly pass &wrapAndCall<fn_type, arg1, arg2, ...> , or wrap it into another lambda which forwards its arguments.

This also means that you cannot overload or specialize wrapAndCallLambda , but you can easily write another overload for wrapAndCall .

You also can never find wrapAndCallLambda via ADL. Which may be a feature you want.