c++ - c++朋友模板函数错误

Question

我可以使用 gcc 6.5 编译以下代码。但是，它无法通过 gcc 5.5 的编译。谁能分享我一些解决方法？

#include <string>
#include <memory>
#include <type_traits>
#include <functional>
#include <utility>
#include <iostream>
#include <experimental/optional>
#include <tuple>

template<typename T>
class Future;
template<typename T, template<typename> class FutureType = Future>
std::shared_ptr<FutureType<T>> MakeReadyFuture(T&& v);

template<typename T>
class Future {
protected:
    template<typename O, template<typename> class FutureType>
    std::shared_ptr<FutureType<O>> friend MakeReadyFuture(O&& v);

    Future(T& v) {}
};

template<typename T, template<typename> class FutureType>
std::shared_ptr<FutureType<T>> MakeReadyFuture(T&& v) {
    return std::shared_ptr<FutureType<T>>(new FutureType<T>(v));
}

int main() {
    MakeReadyFuture(1);
    return 0;
}

错误是：

21:5：错误：'Future::Future(T&) [with T = int]' 受保护

可以在https://godbolt.org/z/ZXlHK1上测试

标签： c++gcc

Answer 1

看起来像 GCC 的错误。这是我在 GCC 5.5 上编译的解决方法：

#include <string>
#include <memory>
#include <type_traits>
#include <functional>
#include <utility>
#include <iostream>
#include <experimental/optional>
#include <tuple>

template<typename T>
class Future;

struct FutureCreator{
    template<typename T, template<typename> class FutureType = Future>
    static std::shared_ptr<FutureType<T>> MakeReadyFuture(T&& v) {
        return std::shared_ptr<FutureType<T>>(new FutureType<T>(v));
    }
};


template<typename T>
class Future {
protected:
    friend FutureCreator;

    Future(T& v) {}
};


int main() {
    FutureCreator::MakeReadyFuture(1);
    return 0;
}