Is there a way to set a template parameter from a variable?

Question

I have an enum value as a member of a class which I want to pass as a template argument? The compiler complains that the member can not be used in a constant expression. Is there any magic to get this work?

My current solution is a switch-case statement, but in my original code EType has nearly 200 entries. So my initial idea was to write a type_traits for mapping the enum values to types.

Here is an examples (also on ideone.com) for my problem (the problem is the last line in main()):

#include <iostream>

enum EType
{
    eType_A,
    eType_B,
    eType_C
};

struct Foo
{
    Foo(EType eType)
        : m_eType(eType)
    {
    }

    EType m_eType;
};

template <EType eType>
struct Bar
{
    static std::string const toString()
    {
        return "-";
    }
};

template <>
struct Bar<eType_A>
{
    static std::string const toString()
    {
        return "A";
    }
};

template <>
struct Bar<eType_B>
{
    static std::string const toString()
    {
        return "B";
    }
};

int main(int argc, char *argv[])
{
    std::cout << "Bar<eType_A>::toString()=" << Bar<eType_A>::toString() << "\n";
    std::cout << "Bar<eType_B>::toString()=" << Bar<eType_B>::toString() << "\n";
    std::cout << "Bar<eType_C>::toString()=" << Bar<eType_C>::toString() << "\n";

    Foo stFooA(eType_A);
    std::cout << "Bar<stFooA.m_eType>::toString()=" << Bar<stFooA.m_eType>::toString() << "\n"; // <--- here ist the problem
}

The examples generates these errors:

prog.cpp: In function ‘int main(int, char**)’:
prog.cpp:54: error: ‘stFooA’ cannot appear in a constant-expression
prog.cpp:54: error: `.' cannot appear in a constant-expression
prog.cpp:54: error: template argument 1 is invalid

greatwolf · Accepted Answer · 2012-01-27 08:06:44Z

5

Template arguments you pass into Bar must be known at compile-time. You're getting that error because stFooA.m_eType can change during run-time so that won't work.

To answer your other question, is there some magic to getting this to work? Perhaps maybe -- are you allowed to make the value of m_eType known at compile-time? If that's a valid constraint for your problem you can change it to something like this and it'll work:

// ...
template <EType eType>
struct Foo
{
    Foo()
    {
    }

    static const EType m_eType = eType;
};

int main(int argc, char *argv[])
{
    std::cout << "Bar<eType_A>::toString()=" << Bar<eType_A>::toString() << "\n";
    std::cout << "Bar<eType_B>::toString()=" << Bar<eType_B>::toString() << "\n";
    std::cout << "Bar<eType_C>::toString()=" << Bar<eType_C>::toString() << "\n";

    Foo<eType_A> stFooA;
    std::cout << "Bar<stFooA.m_eType>::toString()=" 
              << Bar<Foo<eType_A>::m_eType>::toString() 
              << "\n"; // Foo<eType_A>::m_eType is known at compile-time so this works
}

edited Jan 27, 2012 at 8:06

answered Jan 27, 2012 at 7:59

greatwolf

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2 Comments

Lars Over a year ago

Good point, but there is no way to make m_eType known at compile-time.

greatwolf Over a year ago

@Lars in that case then you have no choice -- templates probably isn't the right tool for the task and you'll have to resolve it during run-time somehow. Perhaps turn the template argument into a regular function parameter for toString?

Bo Persson · Accepted Answer · 2012-01-27 07:58:41Z

3

No, the template expansion is done at compile time. It only works if the value is known when the program is compiled.

The compiler also asks for a constant expression.

answered Jan 27, 2012 at 7:58

Bo Persson

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