Is it possible to define a lambda in C++ with default generic argument?

Question

Is it possible to define a lambda in C++ with default generic argument?

int main(){
    auto lambda = [](auto i = 0){return i;};
    std::cout<<lambda()<<"\n"; // this does not compile
    std::cout<<lambda(4)<<"\n"; // this does compile
    auto lambda2 = [](int i = 0){return i;};
    std::cout<<lambda2()<<"\n"; // this is also OK
}

I wonder if it's possible to reproduce something like this functor, and why not

struct Lambda{
    template<typename T=int>
    auto operator()(T i=1){ return i;}
};

Answer 1

You can do that with a wrapper:

template<class F, class DefaultArg>
struct DefaultArgWrapper
{
    F f;
    DefaultArg default_arg;

    template<class... Args>
    decltype(auto) operator()(Args&&... args) {
        return f(std::forward<Args>(args)...);
    }

    decltype(auto) operator()() {
        return f(default_arg);
    }
};

template<class F, class DefaultArg>
DefaultArgWrapper<F, DefaultArg> with_default_arg(F&& f, DefaultArg arg) {
    return {std::move(f), std::move(arg)};
}

int main(){
    auto lambda = with_default_arg([](auto i){return i;}, 0);
    std::cout<<lambda()<<"\n"; 
    std::cout<<lambda(4)<<"\n";
}

---

An alternative C++17 solution:

template<class... F>
struct ComposeF : F... {
    template<class... F2>
    ComposeF(F2&&... fs)
        : F(std::forward<F2>(fs))...
    {}
    using F::operator()...;
};

template<class... F>
ComposeF<std::decay_t<F>...> composef(F&&... fs) {
    return {std::forward<F>(fs)...};
}

int main() {
    auto lambda = [](auto i) { return i; };
    auto f = composef(lambda, [&lambda](int i = 0) { return lambda(i); });
    std::cout << f() << '\n';
    std::cout << f(1) << '\n';
}

A bit sub-optimal though, since there are two copies of lambda involved: one copy in ComposeF, another is the original lambda on the stack. If lambda is mutable that would be an issue.