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Constexpr if testing one template parameter


By : Wilco van der Mersch
Date : September 23 2020, 03:00 AM
I wish did fix the issue. You're too focused on new language features instead of old techniques. If you want a template function that only works if the type given is any instance of MyTemplateClass which takes as its first parameter char, then write that:
code :
template<int32_t N>
void DoStuff(const MyTemplateClass<char, N> &myObj)
{
  // test passed!
}
template<typename ...Args>
void DoStuff(const SomeClass<known, params, ...Args> &myObj);


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How to pass constexpr as a template parameter?


By : EPREVOST
Date : March 29 2020, 07:55 AM
this will help I have a templated class MyClass and I want to run it for various parameters in order to measure some values. I know the exact parameters before the compilation hence I assume there must be a way of achieving the goal.
code :
#include <utility>
#include <cstddef>

constexpr int PARAMS[] = { 1, 2, 3, /*...*/ };

template <int N>
void bar()
{
    MyClass<N> myClass;
    // do sth
}

template <std::size_t... Is>
void foo(std::index_sequence<Is...>)
{
    using dummy = int[];    
    static_cast<void>(dummy{ 0, (bar<PARAMS[Is]>(), 0)... });

    // (bar<PARAMS[Is]>(), ...); since C++1z
}

int main()
{
    foo(std::make_index_sequence<sizeof(PARAMS)/sizeof(*PARAMS)>{});
    //                          <std::size(PARAMS)> since C++1z
    //                          <PARAMS.size()> for std::array<int,N> PARAMS{};
}

Constexpr member function returning a template parameter not considered constexpr when accessed through a reference


By : treg11
Date : March 29 2020, 07:55 AM
I wish this helpful for you Something passed by reference cannot be constexpr. And, for a function to actually be used as constexpr all of the arguments and objects used within the function must also be constexpr. Now, since a reference (even of a template object) can never be constexpr, this means that the better choice is for the compiler to throw an error and let the developer know, rather than the alternative which is to generate the non-constexpr version of the function.
This is because reference means it must have a runtime memory location (see C++11 rvalue and lvalue, and rvalue references). If something qualifies as constexpr it means it is only known and used at compile time, and the result is baked into the resulting executable the same way as hardcoded numbers or strings.

MSVC swallows const from fundamental template parameter in variadic template methods using constexpr if


By : PumpkinRadio Chillou
Date : March 29 2020, 07:55 AM
With these it helps So it turned out to be a bug. For anyone experiencing the same issue, Microsoft said they will have it fixed in VS 15.8 Preview 4.

Using a class template's static constexpr as the template parameter


By : BestinChennai
Date : March 29 2020, 07:55 AM
This might help you I have a base class that contains a buffer with a templated size. I then create subclasses with specific sizes. The sizes should all be multiples of the same number. I thought I'd make that number a constexpr in the base class, but the compiler won't accept the constexpr as a template parameter. Here's some example code: ,
The sizes should all be multiples of the same number.
code :
template<unsigned int Multiplier=1> class A
{
    public:
        static constexpr int basic_size = 256;
    protected:
        int buf[buf_size*Multiplier];
};

class B : A<2>
{
    public:
        void other_stuff();
};

Using constexpr function as template parameter


By : Khalid Ahmed
Date : March 29 2020, 07:55 AM
Hope this helps I am trying to use the result of a constexpr function as a template parameter and cannot figure out how to get it to work. I have the following code: , Your code cannot compile because in:
code :
constexpr auto operator()(const slice &data)
{
    return key<data.size()>(_hash, data.data());
}
template <const slice& data>
constexpr auto operator()()
{
    return key<data.size()>(_hash, data.data());
}
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