I am trying to get a simple example to work to understand how to use std::enable_if. After I read this answer, I thought it shouldn't be too hard to come up with a simple example. I want to use std::enable_if to choose between two member-functions and allow only one of them to be used.
Unfortunately, the following doesn't compile with gcc 4.7 and after hours and hours of trying I am asking you guys what my mistake is.
#include <utility>
#include <iostream>
template< class T >
class Y {
public:
template < typename = typename std::enable_if< true >::type >
T foo() {
return 10;
}
template < typename = typename std::enable_if< false >::type >
T foo() {
return 10;
}
};
int main() {
Y< double > y;
std::cout << y.foo() << std::endl;
}
gcc reports the following problems:
% LANG=C make CXXFLAGS="-std=c++0x" enable_if
g++ -std=c++0x enable_if.cpp -o enable_if
enable_if.cpp:12:65: error: `type' in `struct std::enable_if<false>' does not name a type
enable_if.cpp:13:15: error: `template<class T> template<class> T Y::foo()' cannot be overloaded
enable_if.cpp:9:15: error: with `template<class T> template<class> T Y::foo()'
Why doesn't g++ delete the wrong instantiation for the second member function? According to the standard, std::enable_if< bool, T = void >::type only exists when the boolean template parameter is true. But why doesn't g++ consider this as SFINAE? I think that the overloading error message comes from the problem that g++ doesn't delete the second member function and believes that this should be an overload.
One way to solve this problem, specialization of member functions is to put the specialization into another class, then inherit from that class. You may have to change the order of inheritence to get access to all of the other underlying data but this technique does work.
template< class T, bool condition> struct FooImpl;
template<class T> struct FooImpl<T, true> {
T foo() { return 10; }
};
template<class T> struct FoolImpl<T,false> {
T foo() { return 5; }
};
template< class T >
class Y : public FooImpl<T, boost::is_integer<T> > // whatever your test is goes here.
{
public:
typedef FooImpl<T, boost::is_integer<T> > inherited;
// you will need to use "inherited::" if you want to name any of the
// members of those inherited classes.
};
The disadvantage of this technique is that if you need to test a lot of different things for different member functions you'll have to make a class for each one, and chain it in the inheritence tree. This is true for accessing common data members.
Ex:
template<class T, bool condition> class Goo;
// repeat pattern above.
template<class T, bool condition>
class Foo<T, true> : public Goo<T, boost::test<T> > {
public:
typedef Goo<T, boost::test<T> > inherited:
// etc. etc.
};
I made this short example which also works.
#include <iostream>
#include <type_traits>
class foo;
class bar;
template<class T>
struct is_bar
{
template<class Q = T>
typename std::enable_if<std::is_same<Q, bar>::value, bool>::type check()
{
return true;
}
template<class Q = T>
typename std::enable_if<!std::is_same<Q, bar>::value, bool>::type check()
{
return false;
}
};
int main()
{
is_bar<foo> foo_is_bar;
is_bar<bar> bar_is_bar;
if (!foo_is_bar.check() && bar_is_bar.check())
std::cout << "It works!" << std::endl;
return 0;
}
Comment if you want me to elaborate. I think the code is more or less self-explanatory, but then again I made it so I might be wrong :)
You can see it in action here.
For those late-comers that are looking for a solution that "just works":
#include <utility>
#include <iostream>
template< typename T >
class Y {
template< bool cond, typename U >
using resolvedType = typename std::enable_if< cond, U >::type;
public:
template< typename U = T >
resolvedType< true, U > foo() {
return 11;
}
template< typename U = T >
resolvedType< false, U > foo() {
return 12;
}
};
int main() {
Y< double > y;
std::cout << y.foo() << std::endl;
}
Compile with:
g++ -std=gnu++14 test.cpp
Running gives:
./a.out
11
The boolean needs to depend on the template parameter being deduced. So an easy way to fix is to use a default boolean parameter:
template< class T >
class Y {
public:
template < bool EnableBool = true, typename = typename std::enable_if<( std::is_same<T, double>::value && EnableBool )>::type >
T foo() {
return 10;
}
};
However, this won't work if you want to overload the member function. Instead, its best to use TICK_MEMBER_REQUIRES from the Tick library:
template< class T >
class Y {
public:
TICK_MEMBER_REQUIRES(std::is_same<T, double>::value)
T foo() {
return 10;
}
TICK_MEMBER_REQUIRES(!std::is_same<T, double>::value)
T foo() {
return 10;
}
};
You can also implement your own member requires macro like this(just in case you don't want to use another library):
template<long N>
struct requires_enum
{
enum class type
{
none,
all
};
};
#define MEMBER_REQUIRES(...) \
typename requires_enum<__LINE__>::type PrivateRequiresEnum ## __LINE__ = requires_enum<__LINE__>::type::none, \
class=typename std::enable_if<((PrivateRequiresEnum ## __LINE__ == requires_enum<__LINE__>::type::none) && (__VA_ARGS__))>::type
Here is my minimalist example, using a macro.
Use double brackets enable_if((...)) when using more complex expressions.
template<bool b, std::enable_if_t<b, int> = 0>
using helper_enable_if = int;
#define enable_if(value) typename = helper_enable_if<value>
struct Test
{
template<enable_if(false)>
void run();
}
From this post:
Default template arguments are not part of the signature of a template
But one can do something like this:
#include <iostream>
struct Foo {
template < class T,
class std::enable_if < !std::is_integral<T>::value, int >::type = 0 >
void f(const T& value)
{
std::cout << "Not int" << std::endl;
}
template<class T,
class std::enable_if<std::is_integral<T>::value, int>::type = 0>
void f(const T& value)
{
std::cout << "Int" << std::endl;
}
};
int main()
{
Foo foo;
foo.f(1);
foo.f(1.1);
// Output:
// Int
// Not int
}
Source: Stackoverflow.com