...one of the most highly
regarded and expertly designed C++ library projects in the
world.
— Herb Sutter and Andrei
Alexandrescu, C++
Coding Standards
The enable_if
family of templates
is a set of tools to allow a function template or a class template specialization
to include or exclude itself from a set of matching functions or specializations
based on properties of its template arguments. For example, one can define
function templates that are only enabled for, and thus only match, an arbitrary
set of types defined by a traits class. The enable_if
templates can also be applied to enable class template specializations. Applications
of enable_if
are discussed
in length in [1] and [2].
namespace boost { template <class Cond, class T = void> struct enable_if; template <class Cond, class T = void> struct disable_if; template <class Cond, class T> struct lazy_enable_if; template <class Cond, class T> struct lazy_disable_if; template <bool B, class T = void> struct enable_if_c; template <bool B, class T = void> struct disable_if_c; template <bool B, class T> struct lazy_enable_if_c; template <bool B, class T> struct lazy_disable_if_c; }
Sensible operation of template function overloading in C++ relies on the SFINAE (substitution-failure-is-not-an-error) principle [3]: if an invalid argument or return type is formed during the instantiation of a function template, the instantiation is removed from the overload resolution set instead of causing a compilation error. The following example, taken from [1], demonstrates why this is important:
int negate(int i) { return -i; } template <class F> typename F::result_type negate(const F& f) { return -f(); }
Suppose the compiler encounters the call negate(1)
.
The first definition is obviously a better match, but the compiler must
nevertheless consider (and instantiate the prototypes) of both definitions
to find this out. Instantiating the latter definition with F
as int
would result in:
int::result_type negate(const int&);
where the return type is invalid. If this were an error, adding an unrelated
function template (that was never called) could break otherwise valid code.
Due to the SFINAE principle the above example is not, however, erroneous.
The latter definition of negate
is simply removed from the overload resolution set.
The enable_if
templates
are tools for controlled creation of the SFINAE conditions.
The names of the enable_if
templates have three parts: an optional lazy_
tag, either enable_if
or
disable_if
, and an optional
_c
tag. All eight combinations
of these parts are supported. The meaning of the lazy_
tag is described in the section below.
The second part of the name indicates whether a true condition argument should
enable or disable the current overload. The third part of the name indicates
whether the condition argument is a bool
value (_c
suffix), or a type
containing a static bool
constant
named value
(no suffix).
The latter version interoperates with Boost.MPL.
The definitions of enable_if_c
and enable_if
are as follows
(we use enable_if
templates
unqualified but they are in the boost
namespace).
template <bool B, class T = void> struct enable_if_c { typedef T type; }; template <class T> struct enable_if_c<false, T> {}; template <class Cond, class T = void> struct enable_if : public enable_if_c<Cond::value, T> {};
An instantiation of the enable_if_c
template with the parameter B
as true
contains a member type
type
, defined to be T
. If B
is false
, no such member is
defined. Thus enable_if_c<B,
T>::type
is either a valid or an invalid type
expression, depending on the value of B
.
When valid, enable_if_c<B, T>::type
equals T
. The enable_if_c
template can thus be used for
controlling when functions are considered for overload resolution and when
they are not. For example, the following function is defined for all arithmetic
types (according to the classification of the Boost type_traits
library):
template <class T> typename enable_if_c<boost::is_arithmetic<T>::value, T>::type foo(T t) { return t; }
The disable_if_c
template
is provided as well, and has the same functionality as enable_if_c
except for the negated condition. The following function is enabled for all
non-arithmetic types.
template <class T> typename disable_if_c<boost::is_arithmetic<T>::value, T>::type bar(T t) { return t; }
For easier syntax in some cases and interoperation with Boost.MPL we provide
versions of the enable_if
templates taking any type with a bool
member constant named value
as the condition argument. The MPL bool_
,
and_
, or_
,
and not_
templates are likely
to be useful for creating such types. Also, the traits classes in the Boost.Type_traits
library follow this convention. For example, the above example function
foo
can be alternatively
written as:
template <class T> typename enable_if<boost::is_arithmetic<T>, T>::type foo(T t) { return t; }
The enable_if
templates are
defined in boost/utility/enable_if.hpp
, which is included by boost/utility.hpp
.
With respect to function templates, enable_if
can be used in multiple different ways:
In the previous section, the return type form of enable_if
was shown. As an example of using the form of enable_if
that works via an extra function parameter, the foo
function in the previous section could also be written as:
template <class T> T foo(T t, typename enable_if<boost::is_arithmetic<T> >::type* dummy = 0);
Hence, an extra parameter of type void*
is added, but it is given a default value
to keep the parameter hidden from client code. Note that the second template
argument was not given to enable_if
,
as the default void
gives the
desired behavior.
Which way to write the enabler is largely a matter of taste, but for certain functions, only a subset of the options is possible:
enable_if
must be used either in the return type or in an extra template parameter.
In a compiler which supports C++0x default arguments for function template
parameters, you can enable and disable function templates by adding an
additional template parameter. This approach works in all situations where
you would use either the return type form of enable_if
or the function parameter form, including operators, constructors, variadic
function templates, and even overloaded conversion operations.
As an example:
#include <boost/type_traits/is_arithmetic.hpp> #include <boost/type_traits/is_pointer.hpp> #include <boost/utility/enable_if.hpp> class test { public: // A constructor that works for any argument list of size 10 template< class... T, typename boost::enable_if_c< sizeof...( T ) == 10, int >::type = 0> test( T&&... ); // A conversion operation that can convert to any arithmetic type template< class T, typename boost::enable_if< boost::is_arithmetic< T >, int >::type = 0> operator T() const; // A conversion operation that can convert to any pointer type template< class T, typename boost::enable_if< boost::is_pointer< T >, int >::type = 0> operator T() const; }; int main() { // Works test test_( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ); // Fails as expected test fail_construction( 1, 2, 3, 4, 5 ); // Works by calling the conversion operator enabled for arithmetic types int arithmetic_object = test_; // Works by calling the conversion operator enabled for pointer types int* pointer_object = test_; // Fails as expected struct {} fail_conversion = test_; }
Class template specializations can be enabled or disabled with enable_if
. One extra template parameter
needs to be added for the enabler expressions. This parameter has the default
value void
. For example:
template <class T, class Enable = void> class A { ... }; template <class T> class A<T, typename enable_if<is_integral<T> >::type> { ... }; template <class T> class A<T, typename enable_if<is_float<T> >::type> { ... };
Instantiating A
with any
integral type matches the first specialization, whereas any floating point
type matches the second one. All other types match the primary template.
The condition can be any compile-time boolean expression that depends on
the template arguments of the class. Note that again, the second argument
to enable_if
is not needed;
the default (void
) is the
correct value.
The enable_if_has_type
template is usable this scenario but instead of using a type traits to
enable or disable a specialization, it use a SFINAE context to check for
the existence of a dependent type inside its parameter. For example, the
following structure extracts a dependent value_type
from T if and only if T::value_type
exists.
template <class T, class Enable = void> class value_type_from { typedef T type; }; template <class T> class value_type_from<T, typename enable_if_has_type<typename T::value_type>::type> { typedef typename T::value_type type; };
Once the compiler has examined the enabling conditions and included the function into the overload resolution set, normal C++ overload resolution rules are used to select the best matching function. In particular, there is no ordering between enabling conditions. Function templates with enabling conditions that are not mutually exclusive can lead to ambiguities. For example:
template <class T> typename enable_if<boost::is_integral<T>, void>::type foo(T t) {} template <class T> typename enable_if<boost::is_arithmetic<T>, void>::type foo(T t) {}
All integral types are also arithmetic. Therefore, say, for the call foo(1)
, both
conditions are true and both functions are thus in the overload resolution
set. They are both equally good matches and thus ambiguous. Of course,
more than one enabling condition can be simultaneously true as long as
other arguments disambiguate the functions.
The above discussion applies to using enable_if
in class template partial specializations as well.
In some cases it is necessary to avoid instantiating part of a function signature unless an enabling condition is true. For example:
template <class T, class U> class mult_traits; template <class T, class U> typename enable_if<is_multipliable<T, U>, typename mult_traits<T, U>::type>::type operator*(const T& t, const U& u) { ... }
Assume the class template mult_traits
is a traits class defining the resulting type of a multiplication operator.
The is_multipliable
traits
class specifies for which types to enable the operator. Whenever is_multipliable<A, B>::value
is true
for some types A
and B
,
then mult_traits<A, B>::type
is defined.
Now, trying to invoke (some other overload) of operator*
with, say, operand types C
and D
for which is_multipliable<C, D>::value
is false
and mult_traits<C, D>::type
is not defined is an error on some compilers. The SFINAE principle is not
applied because the invalid type occurs as an argument to another template.
The lazy_enable_if
and
lazy_disable_if
templates
(and their _c
versions)
can be used in such situations:
template<class T, class U> typename lazy_enable_if<is_multipliable<T, U>, mult_traits<T, U> >::type operator*(const T& t, const U& u) { ... }
The second argument of lazy_enable_if
must be a class type that defines a nested type named type
whenever the first parameter (the condition) is true.
Note | |
---|---|
Referring to one member type or static constant in a traits class causes
all of the members (type and static constant) of that specialization
to be instantiated. Therefore, if your traits classes can sometimes contain
invalid types, you should use two distinct templates for describing the
conditions and the type mappings. In the above example, |
Some compilers flag functions as ambiguous if the only distinguishing factor is a different condition in an enabler (even though the functions could never be ambiguous). For example, some compilers (e.g. GCC 3.2) diagnose the following two functions as ambiguous:
template <class T> typename enable_if<boost::is_arithmetic<T>, T>::type foo(T t); template <class T> typename disable_if<boost::is_arithmetic<T>, T>::type foo(T t);
Two workarounds can be applied:
Use an extra dummy parameter which disambiguates the functions. Use a default value for it to hide the parameter from the caller. For example:
template <int> struct dummy { dummy(int) {} }; template <class T> typename enable_if<boost::is_arithmetic<T>, T>::type foo(T t, dummy<0> = 0); template <class T> typename disable_if<boost::is_arithmetic<T>, T>::type foo(T t, dummy<1> = 0);
Define the functions in different namespaces and bring them into a
common namespace with using
declarations:
namespace A { template <class T> typename enable_if<boost::is_arithmetic<T>, T>::type foo(T t); } namespace B { template <class T> typename disable_if<boost::is_arithmetic<T>, T>::type foo(T t); } using A::foo; using B::foo;
Note that the second workaround above cannot be used for member templates. On the other hand, operators do not accept extra arguments, which makes the first workaround unusable. As the net effect, neither of the workarounds are of assistance for templated operators that need to be defined as member functions (assignment and subscript operators).
We are grateful to Howard Hinnant, Jason Shirk, Paul Mensonides, and Richard Smith whose findings have influenced the library.