👀 Views: 58 💬 Answers: 1 📅 Created: 2025-07-14

c++17 stdoptional templates C++

I've been working on this all day and I'm prototyping a solution and Hey everyone, I'm running into an issue that's driving me crazy. I'm having trouble with `std::optional` when trying to use it within a template specialization in C++17. My intention was to provide a default value if the optional is empty, but it seems to be bypassing this logic under certain conditions. Here's a simplified version of what I'm working with: ```cpp #include <iostream> #include <optional> template <typename T> class Wrapper { public: std::optional<T> value; Wrapper(std::optional<T> val) : value(val) {} void printValue() { if (value.has_value()) { std::cout << "Value: " << value.value() << std::endl; } else { std::cout << "Value is empty, using default: " << getDefault() << std::endl; } } T getDefault(); // To be specialized }; template <> int Wrapper<int>::getDefault() { return 42; } int main() { Wrapper<int> w1(std::nullopt); w1.printValue(); // Expecting default value Wrapper<int> w2(10); w2.printValue(); // Should print 10 return 0; } ``` When I run this code, I see the expected output for `w2`, but for `w1`, it prints `Value is empty, using default: 42` instead of using the `getDefault()` method. It seems that the call to `getDefault()` is executed as expected, but the method itself does not seem to be called properly. I've checked the specialization, and it seems correct to me. I also confirmed that the function signature matches the expected type. Can someone help me understand why `getDefault()` isn't behaving as I expected in the context of my `Wrapper` class? I've tried various syntaxes for the specialization, but nothing seems to change the behavior. This is part of a larger application I'm building. Any examples would be super helpful. The stack includes C++ and several other technologies. Cheers for any assistance! I'd be grateful for any help.