What is the argument type deduction rule for template of template?-CodePudding

I have three functions which use template of template:

template <template <typename...> class ContainerType, typename ItemType>
bool has_item(ContainerType<ItemType> items, ItemType target_item)
{
    // ...
}

template <template <typename...> class ContainerType, typename ItemType>
ContainerType<ItemType> filter(ContainerType<ItemType> items, const std::function <bool (ItemType)>& f)
{
   // ...
}


template <template <typename...> class ContainerType, typename ItemType>
bool is_vector(ContainerType<ItemType> items)
{
   // ...
}

I supposed that compiler can deduce the argument types successfully, but it seems that the second one can not be deduced.

    std::vector<int> v = {1, 2, 3, 4, 5};
    std::cout << has_item(v, 1) << std::endl;     // OK
    
    auto less_four = [](int x) { return x < 4; };
    std::vector<int> v2 = filter<std::vector, int>(v, less_four);   // Can not be deduced automatically by compiler
    
    std::cout << is_vector(v2) << std::endl;     // OK

(Here is the Demo)

What is the difference between these three functions, leads to the compiler fail to deduce types automatically?

CodePudding user response：

Because for the 2nd one, template argument deduction for ItemType fails on the 2nd function parameter; implicit conversion (from lambda to std::function) won't be considered in template argument deduction.

You can use std::type_identity (since C 20) to exclude the 2nd function parameter from template argument deduction. E.g.

template <template <typename...> class ContainerType, typename ItemType>
ContainerType<ItemType> filter(ContainerType<ItemType> items, const std::function <bool (std::type_identity_t<ItemType>)>& f)
{
   // ...
}

You can also add another template parameter and stop using std::function.