CODEWITHSUNDEEP
c++templatesc++11enable-if
Jonathan Mee
Jonathan Mee

Reputation: 38909

How Does std::enable_if work?

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(); }

Upvotes: 31

Views: 20836

Answers (2)

Jonathan Mee
Jonathan Mee

Reputation: 38909

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 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 prior to : 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

Upvotes: 31

firda
firda

Reputation: 3338

template<typename T, std::enable_if<std::is_integral<T>::value, int>::type = 0>
void foo(const T& bar) { isInt(); }

this fails to compile if T is not integral (because enable_if<...>::type won't be defined). It is protection of the function foo.The assignment = 0 is there for default template parameter to hide it.

Another possibility: (yes the typename is missing in original question)

#include <type_traits>

template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
void foo(const T& bar) {}

template<typename T>
typename std::enable_if<std::is_integral<T>::value>::type
bar(const T& foo) {}

int main() {
    foo(1); bar(1);
    foo("bad"); bar("bad");
}

error: no matching function for call to ‘foo(const char [4])’
  foo("bad"); bar("bad");
           ^
note: candidate is:
note: template::value, int>::type  > void foo(const T&)
 void foo(const T& bar) {}
      ^
note:   template argument deduction/substitution failed:
error: no type named ‘type’ in ‘struct std::enable_if’
 template::value, int>::type = 0>
                                                                                       ^
note: invalid template non-type parameter
error: no matching function for call to ‘bar(const char [4])’
  foo("bad"); bar("bad");
                       ^

Upvotes: 10

Related Questions