Identifying primitive types in templates

Question

Hello

I am looking for a way to identify primitives types in a template class definition.

I mean, having this class :

template<class T>
class A{
void doWork(){
   if(T isPrimitiveType())
     doSomething();
   else
     doSomethingElse(); 
}
private:
T *t; 
};

Is there is any way to "implement" isPrimitiveType().

Answer 1

// Generic: Not primitive
template<class T>
bool isPrimitiveType() {
    return false;
}

// Now, you have to create specializations for **all** primitive types

template<>
bool isPrimitiveType<int>() {
    return true;
}

// TODO: bool, double, char, ....

// Usage:
template<class T>
void test() {
    if (isPrimitiveType<T>()) {
        std::cout << "Primitive" << std::endl;
    } else {
        std::cout << "Not primitive" << std::endl;
    }
 }

In order to save the function call overhead, use structs:

template<class T>
struct IsPrimitiveType {
    enum { VALUE = 0 };
};

template<>
struct IsPrimitiveType<int> {
    enum { VALUE = 1 };
};

// ...

template<class T>
void test() {
    if (IsPrimitiveType<T>::VALUE) {
        // ...
    } else {
        // ...
    }
}

As others have pointed out, you can save your time implementing that by yourself and use is_fundamental from the Boost Type Traits Library, which seems to do exactly the same.

Answer 2

Boost TypeTraits has plenty of stuff.

Answer 3

Assuming by 'Primitive Type' you mean the built in types you can do a series of template specializations. Your code would become:

template<class T>
struct A{
    void doWork();
private:
    T *t; 
};

template<> void A<float>::doWork()
{
    doSomething();
}

template<> void A<int>::doWork()
{
    doSomething();
}

// etc. for whatever types you like

template<class T> void A<T>::doWork()
{
    doSomethingElse();
}

Answer 4

The following example (first posted in comp.lang.c++.moderated) illustrates using partial specialisation to print things differently depending on whether or not they are built-in types.

// some template stuff
//--------------------
#include <iostream>
#include <vector>
#include <list>

using namespace std;

// test for numeric types
//-------------------------
template <typename T> struct IsNum {
    enum { Yes = 0, No = 1 };
};


template <> struct IsNum <int> { 
    enum { Yes = 1, No = 0 };
};


template <> struct IsNum <double> {
    enum { Yes = 1, No = 0 };
};

// add more IsNum types as required

// template with specialisation for collections and numeric types
//---------------------------------------------------------------
template <typename T, bool num = false> struct Printer {
    void Print( const T & t ) {
        typename T::const_iterator it = t.begin();
        while( it != t.end() ) {
            cout << *it << " ";
            ++it;
        }
        cout << endl;
    }
};

template <typename T> struct Printer <T, true> {
    void Print( const T & t ) {
        cout << t << endl;
    }
};

// print function instantiates printer depoending on whether or
// not we are trying to print numeric type
//-------------------------------------------------------------
template <class T> void MyPrint( const T & t ) {
    Printer <T, IsNum<T>::Yes> p;
    p.Print( t );
}

// some test types
//----------------
typedef std::vector <int> Vec;
typedef std::list <int> List;

// test it all
//------------
int main() {

    Vec x;
    x.push_back( 1 );
    x.push_back( 2 );
    MyPrint( x );        // prints 1 2

    List y;
    y.push_back( 3 );
    y.push_back( 4 );
    MyPrint( y );        // prints 3 4

    int z = 42;
    MyPrint( z );        // prints 42

    return 0;
}

Answer 5

It cannot be done exactly the way you asked. Here is how it can be done :

template<class T>
class A{
void doWork(){
   bool isPrimitive = boost::is_fundamental<T>::value;
   if(isPrimitive)
     doSomething();
   else
     doSomethingElse(); 
}
private:
T *t; 
};

You may get a warning if you put the value of isPrimitive directly inside the if statement. This is why i introduced a temporary variable.

Answer 6

Yet another similar examples:

#include  <boost/type_traits/is_fundamental.hpp>
#include <iostream>

template<typename T, bool=true>
struct foo_impl
{
    void do_work()
    {
        std::cout << "0" << std::endl;
    }
};
template<typename T>
struct foo_impl<T,false>
{
    void do_work()
    {
        std::cout << "1" << std::endl;
    }
};

template<class T>
struct foo
{
    void do_work()
    {
        foo_impl<T, boost::is_fundamental<T>::value>().do_work();
    }
};


int main()
{
    foo<int> a; a.do_work();
    foo<std::string> b; b.do_work();
}