Detecting array type do not works

Question

template<typename T> 
class CompoundT {           // primary template 
  public: 
    enum { IsPtrT = 0, IsRefT = 0, IsArrayT = 0, 
           IsFuncT = 0, IsPtrMemT = 0 }; 
    typedef T BaseT; 
    typedef T BottomT; 
    typedef CompoundT<void> ClassT; 
};

template<typename T, size_t N> 
class CompoundT <T[N]> {    // partial specialization for arrays 
  public: 
    enum { IsPtrT = 0, IsRefT = 0, IsArrayT = 1, 
           IsFuncT = 0, IsPtrMemT = 0 }; 
    typedef T BaseT; 
    typedef typename CompoundT<T>::BottomT BottomT; 
    typedef CompoundT<void> ClassT; 
};

and in main:

template<class T>
bool isArray(T a)
{
    return CompoundT<T>::IsArrayT;
}

int _tmain(int argc, _TCHAR* argv[])
    {
        int a[10];
        cout << isArray(a);
        return 0;
}

Why this doesn't work? This example is from "Templates the complete guide" ch.19.2.

Won't I need size_t N in the first template too so that the second one is considered as a partial specialization of the first? — Arun
– Arun, Commented Oct 10, 2010 at 19:19

Armen Tsirunyan · Accepted Answer · 2010-10-10 19:16:50Z

5

Because isArray must take reference, otherwise if you take an array by value it's the same as if you take a pointer :)

template <class T>
bool isArray(const T& ) {...}

because

void f(int a[10]);

and

void f(int* a);

are equivalent declarations.

answered Oct 10, 2010 at 19:16

Armen Tsirunyan

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3 Comments

Armen Tsirunyan Over a year ago

@There: No, as a matter of fact, I just got home 5 min ago. Was my answer helpful?

There is nothing we can do Over a year ago

yes, thank you. Can't accept though. Have to wait 15 minutes.

Armen Tsirunyan Over a year ago

@There The same is with functions, if you take a function by value, it's the same as if you took a pointer to function. The same is with templates. If a templated function takes T by value and you pass a function then T is a pointer to function. if you take T by reference, T will be function

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