无法推断功能参数的非类型模板参数

Question

This might be something really simple, but I cannot figure out why this code does not compile

#include <type_traits>

template <std::size_t Size, std::size_t Align>
void foo(std::aligned_storage_t<Size, Align>&) {}

int main() {
    auto storage = std::aligned_storage_t<100, 8>{};
    foo(storage);
}

( https://wandbox.org/permlink/PdBwAWVh6N9rkTE2 )

How can I get this to work? And why does not this compile?

---

The usecase is that foo() is a suite of overloads on unrelated types like aligned_storage_t , int , double , etc. And the aligned_storage_t instance represents memory that foo() knows how to reinterpret and use.

Answer 1

It doesn't work because std::aligned_storage_t is not a class, it's a type alias for some implementation defined type. In fact what you have is:

template <std::size_t Size, std::size_t Align>
void foo(typename std::aligned_storage<Size, Align>::type);

Size and Align cannot be deduced from this as it is a non-deduced context. You need to replace both instances of std::aligned_storage_t with std::aligned_storage . Then if you need the aligned type you would access that with ::type .

Answer 2

Size and Align is not deducible with std::aligned_storage<Size, Align>::type .

You might use sizeof / alignof to retrieve (nearly) initial value:

template <typename T>
void foo(const T&)
{
    constexpr std::size_t size = sizeof(T);
    constexpr std::size_t alignment = alignof(T);

    std::cout << size << " "<< alignment << std::endl;
}

Demo (got size of 104 instead of input 100 due to alignment for your example)