I use a template parameter to determine if a certain behavior must be done or not. But this code generate a warning on VS2008 : Warning 26 warning C4127: conditional expression is constant
Here an exemple of the code :
template <class param, bool param2=true>
class superclass1
{
public:
int foo()
{
if(param2)
doSomthingMore();
return 1;
}
};
Is there a way to tranform the code to remove the warning and get the same features?
This is done via partial specialization. The crudest version looks like this:
template <typename, bool> class superclass1;
template <class param> class superclass1<param, true>
class superclass1
{
public:
int foo()
{
doSomthingMore();
return 1;
}
};
template <class param> class superclass1<param, false>
class superclass1
{
public:
int foo()
{
return 1;
}
};
A more sophisticated approach might declare a member template function and only specialize that. Here's a solution with auxiliary tag classes:
#include <type_traits>
template <bool B> class Foo
{
struct true_tag {};
struct false_tag {};
void f_impl(true_tag = true_tag()){} // your code here...
void f_impl(false_tag = false_tag()){} // ... and here
public:
void foo()
{
f(typename std::conditional<B, true_tag, false_tag>::type());
}
};
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