C++ template lambda wrapper

Question

Can anyone figure out how to make this compile ?

I'm trying to wrap a lambda in another function that does something (here printing "you know what") + calling the lambda.

Best would be to have automatic template parameters deduction.

#include <iostream>
#include <functional>
#include <utility>

void youKnowWhat(const std::function<void()>&& fun)
{
    std::cout << "You know what ?" << std::endl;
    fun();
}

template <typename... Args>
auto youKnowWhatSomething(const std::function<void(Args...)>&& fun)
{
    return [fun{std::move(fun)}](Args... args)
        {
            youKnowWhat(std::bind(fun, std::forward<Args>(args)...));
        };
}

int main()
{
    const auto imHavingSomething([](const std::string& s){
        std::cout << "Im having " << s << std::endl;
    });

    const auto youKnowWhatImHavingSomething(youKnowWhatSomething(std::move(imHavingSomething)));
    youKnowWhatImHavingSomething("fun with templates");
    youKnowWhatImHavingSomething("headaches");
}

See [Why cannot a C++ compiler deduce the type of the lambda argument?](https://stackoverflow.com/questions/65060766/why-cannot-a-c-compiler-deduce-the-type-of-the-lambda-argument). — rustyx, Mar 04 '22 at 10:54
`std::move` a const variable is.. a copy (ignoring some exceptions). — Jarod42, Mar 04 '22 at 11:25

score 1 · Answer 1 · answered Mar 04 '22 at 10:34

Will this work for you? I've (basically) removed the const from const ... &&, and added a template parameter to the function youKnowWhatSomething. The lambda is no longer declared as const in main.

#include <iostream>
#include <functional>
#include <utility>

void youKnowWhat(const std::function<void()>& fun)
{
    std::cout << "You know what ?" << std::endl;
    fun();
}

template<typename function_t, typename... Args>
auto youKnowWhatSomething(function_t&& fun)
{
    return [fun{std::move(fun)}](Args... args)
        {
            youKnowWhat(std::bind(fun, std::forward<Args>(args)...));
        };
}

int main() {
    auto imHavingSomething = [](std::string s) {
        std::cout << "Im having " << s << std::endl;
    };

    const auto youKnowWhatImHavingSomething =
        youKnowWhatSomething<decltype(imHavingSomething), std::string>
        (std::move(imHavingSomething));

    youKnowWhatImHavingSomething("fun with templates");
    youKnowWhatImHavingSomething("headaches");
}

Nice :) Do you have a solution with automatic template parameters deduction ? — Adrien Grosjean, Mar 04 '22 at 11:35

Jarod42 · Accepted Answer · 2022-03-24T12:43:38.053

1

How about

#include <iostream>
#include <functional>
#include <utility>

template <typename F>
void youKnowWhat(F&& fun)
{
    std::cout << "You know what ?" << std::endl;
    fun();
}

template <typename F>
auto youKnowWhatSomething(F&& fun)
{
    return [fun{std::move(fun)}](auto&&... args) -> decltype(fun(std::forward<decltype(args)>(args)...), void())
        {
            youKnowWhat([&](){fun(std::forward<decltype(args)>(args)...); });
        };
}

int main()
{
    const auto imHavingSomething([](std::string s){
        std::cout << "Im having " << s << std::endl;
    });

    const auto youKnowWhatImHavingSomething(youKnowWhatSomething(imHavingSomething));
    youKnowWhatImHavingSomething("fun with templates");
    youKnowWhatImHavingSomething("headaches");
}

Demo

edited Mar 24 '22 at 12:43

answered Mar 04 '22 at 12:24

Jarod42

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Thanks ! I changed the code just to keep using std::bind inside, it also works. I don't understand what is the `void` doing in `decltype(fun(std::forward(args)...), void())` ? – Adrien Grosjean Mar 04 '22 at 13:52
1

I use `decltype` for SFINAE, last `void()` is to tell return type is `void`. You might drop the trailling return type for simplicity if wanted. – Jarod42 Mar 04 '22 at 13:58
I came across an issue with the above version : `auto... args` should be `auto&&... args` otherwise if the wrapped function takes a reference it makes a copy instead – Adrien Grosjean Mar 24 '22 at 09:52
1

@AdrienGrosjean: Version improved. – Jarod42 Mar 24 '22 at 12:44

C++ template lambda wrapper

2 Answers2