SFINAE 未检测到 T::reference

发布于 2024-11-04 01:55:26 字数 1194 浏览 0 评论 0 原文

std::vector 类是 STL 容器概念的模型，因此任何正确的向量实现都必须包含嵌套的 typedef value_type 以及参考。这应该可以使用 SFINAE 检测到。不过，在我自己的测试中，我可以使用 SFINAE 来检测嵌套的 value_type typedef，但由于某种原因，我无法检测引用。

template <class T> 
typename T::value_type* test(T)
{
    cout << "Has nested typedef!" << endl;
}

template <class T> 
void test(...)
{
    cout << "Doesn't have nested typedef!" << endl;
}

int main()
{
    test(std::vector<int>());
}

输出： Hasnested typedef!

但是，如果我将 value_type 替换为 reference，例如：

template <class T> 
typename T::reference* test(T)
{
    cout << "Has nested typedef!" << endl;
}

template <class T> 
void test(...)
{
    cout << "Doesn't have nested typedef!" << endl;
}

int main()
{
    test(std::vector<int>());
}

...程序根本无法编译，给出错误： error: nomatching function for call to test(std::vector>)

为什么 SFINAE 技术适用于 T: :value_type 但不与 T::reference 一起使用？

原文

The std::vector<T> class is a model of the STL Container concept, and as such any proper implementation of vector has to include a nested typedef value_type as well as reference. This should be detectable using SFINAE. However, in my own tests, I can use SFINAE to detect a nested value_type typedef, but for some reason I can't detect reference.

template <class T> 
typename T::value_type* test(T)
{
    cout << "Has nested typedef!" << endl;
}

template <class T> 
void test(...)
{
    cout << "Doesn't have nested typedef!" << endl;
}

int main()
{
    test(std::vector<int>());
}

This outputs: Has nested typedef!

However, if I replace value_type with reference, like:

template <class T> 
typename T::reference* test(T)
{
    cout << "Has nested typedef!" << endl;
}

template <class T> 
void test(...)
{
    cout << "Doesn't have nested typedef!" << endl;
}

int main()
{
    test(std::vector<int>());
}

...the program fails to compile at all, giving the error: error: no matching function for call to test(std::vector<int, std::allocator<int> >)

Why does the SFINAE technique work with T::value_type but not with T::reference?