Question

我知道如何pass template function as template argument，我现在正在努力以类似的方式传递变量模板。

这是我尝试过的最小例子：

#define PASS_VARIABLE_TEMPLATE(name) [dummy=nullptr](auto&&...args) \
                                                    {return name<decltype(args)...>;}

//testing
template <typename T>
bool value = std::is_fundamental<T>::value;

template <typename Hax>
void print_bool(Hax h)
{
    std::cout << h(int{}) << std::endl; // no error, wrong output
    //std::cout << h(int{}, float{}) << std::endl; // error, good
}

int main()
{
    print_bool(PASS_VARIABLE_TEMPLATE(value)); //prints 0 instead of 1
}

Demo

如果编译，那么输出错误的原因是什么？

Answer 1

The main problem with your code is that decltype deduces the arguments as an rvalue reference (int&&) because your lambda uses forwarding references to accept the arguments. std::is_fundamental will work well with a bare type.

For your specific snippet, the correct solution is to remove the reference.

#define PASS_VARIABLE_TEMPLATE(name) \
    [dummy=nullptr](auto&&...args){return name<std::remove_reference_t<decltype(args)>...>;}

Now it works. :-) See it Live On Coliru

A slightly more or better generic way will be to additionally remove cv qualifiers. In the end, you may want to use std::decay

#define PASS_VARIABLE_TEMPLATE(name) [dummy=nullptr](auto&&...args) \
{return name<std::decay_t<decltype(args)>...>;}

Answer 2

template<class T>struct tag_t{using type=T; constexpr tag_t(){}};
template<class Tag>using tagged_type=typename Tag::type;
template<class T>constexpr tag_t<T> tag{};

这些帮助将类型作为值传递并解压缩。

#define PASS_VARIABLE_TEMPLATE(name) [](auto...args) \
                                                {return name<tagged_type<decltype(args)>...>;}

在print_bool内进行：

std::cout << h(tag<int>) << std::endl;

不确定为什么要执行dummy=nullptr事。

tag作为模板可以进行未受干扰的类型。

是否可以通过lambda传递变量模板？

2 个答案: