static_assert to check value in non-template class's template constructor

Question

So, I'm looking for a way to cause a compile-time error if the value used in declaring the object is equal to another value (don't wish to use C's assert macro).

Yes, I know why the problem occurs... The compiler was quite clear when he/she complained expression did not evaluate to a constant.

I also don't want to make my entire class a template. Is there a little miracle workaround that I'm missing?

#include <iostream>

class test
{
private:
    template<class T, class T2>
    using both_int = std::enable_if_t<std::is_integral<T>::value && std::is_integral<T2>::value>;
private:
    int _a, _b;
public:

    template<class T, class T2 = T, class = both_int<T, T2>>
    constexpr test(const T &a, const T2 &b = 1)
        : _a(a), _b(b)
    {
        static_assert(b != 0, "Division by zero!");
    }
};

int main()
{
    test obj(1, 0); //ERROR

    test obj2(0, 1); //OK

    std::cin.get();
}

Make b a template parameter of the constructor.

Christian Hackl
– Christian Hackl

2015-11-01 14:56:01 +00:00
Commented Nov 1, 2015 at 14:56 — Christian Hackl
– Christian Hackl, Commented Nov 1, 2015 at 14:56

Community · Accepted Answer · 2017-05-23 11:50:34Z

2

You can do it like this:

struct test {
private:
  constexpr test(int a, int b) {}
public:
  template<int b>
  static test construct(int a) {
    static_assert(b != 0, "not zero failed");
    return test(a, b);
  }     
};

int main()
{
   test foo = test::construct<1>(1);
   test foo = test::construct<0>(1);//compile error
}

You need static constructor, because of there is no way to specify parameters of template constructor, see C++ template constructor

edited May 23, 2017 at 11:50

CommunityBot

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answered Nov 1, 2015 at 15:05

fghj

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