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Home/ Questions/Q 441323
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Editorial Team
  • 0
Editorial Team
Asked: May 12, 20262026-05-12T20:57:40+00:00 2026-05-12T20:57:40+00:00

I want to be able to make the compiler shout when i call a

  • 0

I want to be able to make the compiler shout when i call a constructor of foo with a class
that is NOT derived from _base*. The current code allows only for foo<_base*> itself. Any
easy solution ?

class _base
{
public:
    // ...
};

class _derived: public _base
{
public:
    // ...
};

template <typename T>
class foo
{
public:
    foo ()      { void TEMPLATE_ERROR; }
};

template <> foo<_base*>::foo () 
{
    // this is the only constructor
}

main-code:

foo<_base*>    a;    // should work 
foo<_derived*> b;    // should work (but doesnt)
foo<int*>      c;    // should not work (and infact doesnt)
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T20:57:40+00:00Added an answer on May 12, 2026 at 8:57 pm

    Without Boost you can use something like the following to determine whether a pointer-to-type can be implicitly cast to another pointer-to-type:

    template <class Derived, class Base>
    struct IsConvertible
    {
        template <class T>
        static char test(T*);
    
        template <class T>
        static double test(...);
    
        static const bool value = sizeof(test<Base>(static_cast<Derived*>(0))) == 1;
    };
    

    To make it trigger an error at compile-time, you can now use value in an expression that causes an error if it is false, for example typedef a negative-sized array.

    template <typename T>
    class foo
    {
    public:
        foo ()
        {
            typedef T assert_at_compile_time[IsConvertible<T, _base>::value ? 1 : -1];
        }
    };
    
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