c++ - 如何使用非平凡的析构函数防止未使用的变量警告

Question

当我依靠生命周期扩展来分配给具有非平凡析构函数的类时，编译器(gcc 和 clang)会发出未使用的变量警告。有没有办法解决这个问题？ https://wandbox.org/permlink/qURr4oliu90xJpqr

#include <iostream>

using std::cout;
using std::endl;

class Something {
public:
    explicit Something(int a_in) : a{a_in} {}

    Something() = delete;
    Something(Something&&) = delete;
    Something(const Something&) = delete;
    Something& operator=(Something&&) = delete;
    Something& operator=(const Something&) = delete;

    ~Something() {
        cout << this->a << endl;
    }

private:
    int a;
};

int main() {
    const auto& something = Something{1};
    return 0;
}

请注意，当我切换到不依赖生命周期延长时，一切正常 https://wandbox.org/permlink/cJFwUDdi1YUEWllq

我什至尝试手动定义所有构造函数，然后使用模板化的 static_assert 删除它们，这样它只会在调用这些构造函数时触发 https://wandbox.org/permlink/fjHJRKG9YW6VGOFb

#include <iostream>

using std::cout;
using std::endl;

template <typename Type>
constexpr auto definitely_false = false;

template <typename T = void>
class Something {
public:
    explicit Something(int a_in) : a{a_in} {}

    Something() { static_assert(definitely_false<T>, ""); }
    Something(Something&&) { static_assert(definitely_false<T>, ""); }
    Something(const Something&) { static_assert(definitely_false<T>, ""); }
    Something& operator=(Something&&) { static_assert(definitely_false<T>, ""); }
    Something& operator=(const Something&) { static_assert(definitely_false<T>, ""); }

    ~Something() {
        cout << this->a << endl;
    }

private:
    int a;
};

int main() {
    const auto& something = Something<>{1};
    return 0;
}

为了语言律师标签，我们假设强制转换为 void 不是一个选项。我可以对构造函数/析构函数做些什么来帮助消除此警告吗？

Answer 1

不是您正在寻找的确切答案，但这里有一个解决方法建议:

C++17 前

使用 std::ignore像这样:

const auto& something = Something{1};
std::ignore = something;

---

后 C++17

使用 maybe_unused attibute ，像这样:

[[maybe_unused]] const auto& something = Something{1};