Re: template type only known at runtime

From:
Leandro Melo <ltcmelo@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 18 Nov 2008 03:19:44 -0800 (PST)
Message-ID:
<f82e9676-805c-4cfa-a6d4-3c72df4ad14a@s9g2000prg.googlegroups.com>
On 18 nov, 08:51, Ralf Goertz
<r_goe...@expires-2006-11-30.arcornews.de> wrote:

Hi,

I want to do able to declare an object bar of a templated class foo
where the actual template type of foo is only know at runtime:

-----

#include <string>

template <class T> class foo {
    T x;
public:
    void do_something_depending_on_type_of_T(){

        ...
    };

}

int main(int argc, char *argv[]) {
    if (argc==1) //default type string should be used
        foo<std::string> bar;
    else //use class int
        foo<int> bar;
    bar.do_something_depending_on_type_of_T();
    return 0;

}


Template based code is constructed at compile time (static
polymorphism). If you need dynamic polymorphism use virtual functions.
However, your code has a general C++ problem (not specific to
templates). The line bar.do_something_depending_on_type_of_T(); will
generate an error because bar is undefined. This would happen even if
bar was not a template.

I don't know if this is the best design for you. But the closest I can
get to it is through template specialization:

emplate <class T> class foo
{
public:
    void do_something()
    {} //You don't need trailing semi-colon here.
}; //You need it here though.

template <> class foo<std::string>
{
public:
    void do_something()
    { /*Implementation for string */ }
};

template <> class foo<int>
{
public:
    void do_something()
    { /*Implementation for int */ }
};

int main(int argc, char *argv[])
{
    foo<std::string> stringBar;
    foo<int> intBar;
    if (argc==1)
        stringBar.do_something();
    else
        intBar.do_something();
    return 0;
}

--
Leandro T. C. Melo

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