Reputation: 703
I looked at a few similar questions, like this one and this other one, for example, and I understand how to work with enable_if for member functions.
Here is a working example:
#include <iostream>
template <int size>
class Test
{
private:
constexpr static bool ENABLE = (size < 10);
public:
template <bool E = ENABLE, typename std::enable_if<E, int>::type = 0>
static int foo();
template <bool E = ENABLE, typename std::enable_if<!E, int>::type = 0>
constexpr static int foo();
};
template <int size>
template <bool E, typename std::enable_if<E, int>::type>
int Test<size>::foo()
{
return 7;
}
template <int size>
template <bool E, typename std::enable_if<!E, int>::type>
constexpr int Test<size>::foo()
{
return 12;
}
int main()
{
Test<5> v1;
Test<15> v2;
std::cout << v1.foo() << "\n";
std::cout << v2.foo() << "\n";
}
However, when I try to slightly modify the code to work for member variables, I get nasty redeclaration errors. Is this even possible to do with variables, am I just missing something simple?
Here is my problematic example code:
#include <iostream>
template <int size>
class Test
{
private:
constexpr static bool ENABLE = (size < 10);
public:
template <bool E = ENABLE, typename std::enable_if<E, int>::type = 0>
static int foo;
template <bool E = ENABLE, typename std::enable_if<!E, int>::type = 0>
constexpr static int foo = 12;
};
template <int size>
template <bool E, typename std::enable_if<E, int>::type>
int Test<size>::foo = 7;
template <int size>
template <bool E, typename std::enable_if<!E, int>::type>
constexpr int Test<size>::foo;
int main()
{
Test<5> v1;
Test<15> v2;
std::cout << v1.foo<> << "\n";
std::cout << v2.foo<> << "\n";
}
Thanks in advance, any help/guidance is appreciated!
Upvotes: 3
Views: 1140
Reputation: 4896
If you are only doing this on static member variables, there is still a chance you can make them work, by using block scoped static variables. This way your foo
becomes a function, so you can apply SFINAE on them.
#include <iostream>
template <int size>
class Test
{
private:
constexpr static bool ENABLE = (size < 10);
public:
template <bool E = ENABLE, typename std::enable_if<E, int>::type = 0>
static int& foo();
template <bool E = ENABLE, typename std::enable_if<!E, int>::type = 0>
constexpr static int foo();
};
template <int size>
template <bool E, typename std::enable_if<E, int>::type>
int& Test<size>::foo() {
static int foo_impl = 7;
return foo_impl;
}
template <int size>
template <bool E, typename std::enable_if<!E, int>::type>
constexpr int Test<size>::foo() {
return 12;
}
int main()
{
Test<5> v1;
Test<15> v2;
std::cout << v1.foo() << "\n";
std::cout << v2.foo() << "\n";
}
Upvotes: 0
Reputation: 136515
You can achieve the desired effect with a conditional base class that provides member foo
:
template<int FOO_INIT>
struct TestImpl1 {
static int foo;
};
template<int FOO_INIT>
int TestImpl1<FOO_INIT>::foo = FOO_INIT;
template<int FOO_INIT>
struct TestImpl2 {
constexpr static int foo = FOO_INIT;
};
template<int FOO_INIT>
constexpr int TestImpl2<FOO_INIT>::foo;
template<int size>
struct Test
: std::conditional<
(size < 10),
TestImpl1<7>,
TestImpl2<12>
>::type
{};
int main() {
Test<5> v1;
Test<15> v2;
std::cout << v1.foo << "\n";
std::cout << v2.foo << "\n";
// constexpr int i1 = v1.foo; // Fails to compile because Test<5>::foo is not constexpr.
constexpr int i2 = v2.foo; // Compiles because Test<15>::foo is constexpr.
}
Upvotes: 6