I just asked this question: std::numeric_limits as a Condition
I understand the usage where std::enable_if
will define the return type of a method conditionally causing the method to fail to compile.
template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }
What I don't understand is the second argument and the seemingly meaningless assignment to std::enable_if
when it's declared as part of the template statement, as in Rapptz answer.
template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
void foo(const T& bar) { isInt(); }
As is mentioned in comment by 40two, understanding of Substitution Failure Is Not An Error is a prerequisite for understanding std::enable_if
.
std::enable_if
is a specialized template defined as:
template<bool Cond, class T = void> struct enable_if {};
template<class T> struct enable_if<true, T> { typedef T type; };
The key here is in the fact that typedef T type
is only defined when bool Cond
is true
.
Now armed with that understanding of std::enable_if
it's clear that void foo(const T &bar) { isInt(bar); }
is defined by:
template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }
As mentioned in firda's answer, the = 0
is a defaulting of the second template parameter. The reason for the defaulting in template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
is so that both options can be called with foo< int >( 1 );
. If the std::enable_if
template parameter was not defaulted, calling foo
would require two template parameters, not just the int
.
General note, this answer is made clearer by explicitly typing out typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type
but void
is the default second parameter to std::enable_if
, and if you have c++14 enable_if_t
is a defined type and should be used. So the return type should condense to: std::enable_if_t<std::numeric_limits<T>::is_integer>
A special note for users of visual-studio prior to visual-studio-2013: Default template parameters aren't supported, so you'll only be able to use the enable_if
on the function return: std::numeric_limits as a Condition