C++14 Lambda - 通过引用或值有条件地捕获

Question

是否可以根据编译时信息有条件地选择lambda的捕获方法？例如……

auto monad = [](auto && captive) {
    return [(?)captive = std::forward<decltype(captive)>(captive)](auto && a) {
        return 1;
    };
};

如果 decltype(captive) 是一个 std::reference_wrapper，我想通过引用捕获，而其他一切都通过值捕获。

Answer 1

Lambda 捕获类型不能由依赖于模板的名称控制。

但是，您可以通过将创建内部 lambda 委托(delegate)给重载函数来达到预期的效果:

template<class T>
auto make_monad(T&& arg) {
    return [captive = std::forward<T>(arg)](auto&& a) {
        std::cout << __PRETTY_FUNCTION__ << " " << a << '\n';
        return 1;
    };
}

template<class T>
auto make_monad(std::reference_wrapper<T> arg) {
    return [&captive = static_cast<T&>(arg)](auto&& a) {
        std::cout << __PRETTY_FUNCTION__ << " " << a << '\n';
        return 1;
    };
}

int main() {
    auto monad = [](auto&& captive) {
        return make_monad(std::forward<decltype(captive)>(captive));
    };

    int n = 1;
    monad(1)(1);
    monad(n)(2);
    monad(std::ref(n))(3);
}

输出:

make_monad(T&&)::<lambda(auto:1&&)> [with auto:1 = int; T = int] 1
make_monad(T&&)::<lambda(auto:1&&)> [with auto:1 = int; T = int&] 2
make_monad(std::reference_wrapper<_Tp>)::<lambda(auto:2&&)> [with auto:2 = int; T = int] 3

---

I don't want to capture reference_wrapper by reference, I want to capture the reference it holds by reference. Reference wrapper does it's best to be a like a reference, but since the call operator (aka, "." operator) cannot be overloaded, it fails pretty miserably at the end of the day.

在这种情况下，您不需要更改 std::reference_wrapper<T> 的捕获类型.相反，您可能希望像任何其他类型的参数一样按值捕获它，并在使用站点首先解开参数:

template<class T> T& unwrap(T& t) { return t; }
template<class T> T& unwrap(std::reference_wrapper<T> t) { return t; }

auto monad = [](auto && captive) {
    return [captive](auto && a) {            // <--- Capture by value.
        auto& captive_ref = unwrap(captive); // <--- Unwrap before usage.
        return 1;
    };
};