I have the following case that works using std::enable_if
:
template<typename T, typename std::enable_if<std::is_same<int, T>::value>::type* = nullptr> void f() { } template<typename T, typename std::enable_if<std::is_same<double, T>::value>::type* = nullptr> void f() { }
Now, I saw in cppreference the new syntax, much cleaner in my opinion : typename = std::enable_if_t<std::is_same<int, T>::value>>
I wanted to port my code :
template<typename T, typename = std::enable_if_t<std::is_same<int, T>::value>> void g() { } template<typename T, typename = std::enable_if_t<std::is_same<double, T>::value>> void g() { }
But now GCC (5.2) complains :
error: redefinition of 'template<class T, class> void g()' void g() { }
Why is that so ? What can I do to have the new, more concise syntax in this case if this is possible ?
A template parameter is a special kind of parameter that can be used to pass a type as argument: just like regular function parameters can be used to pass values to a function, template parameters allow to pass also types to a function.
1) A template template parameter with an optional name. 2) A template template parameter with an optional name and a default. 3) A template template parameter pack with an optional name.
" typename " is a keyword in the C++ programming language used when writing templates. It is used for specifying that a dependent name in a template definition or declaration is a type.
Let's remove some code.
template< class T, class U/* = std::enable_if_t<std::is_same<int, T>::value>*/ > void g() { } template< class T, class U/* = std::enable_if_t<std::is_same<double, T>::value>*/ > void g() { }
would you be surprised if the compiler rejected the two above templates?
They are both template functions of "type" template<class,class>void()
. The fact that the 2nd type argument has a different default value matters not. That would be like expecting two different print(string, int)
functions with different default int
values to overload. ;)
In the first case we have:
template< typename T, typename std::enable_if<std::is_same<int, T>::value>::type* = nullptr > void f() { } template< typename T, typename std::enable_if<std::is_same<double, T>::value>::type* = nullptr > void f() { }
here we cannot remove the enable_if
clause. Updating to enable_if_t
:
template< class T, std::enable_if_t<std::is_same<int, T>::value, int>* = nullptr > void f() { } template< class T, std::enable_if_t<std::is_same<double, T>::value, int>* = nullptr > void f() { }
I also replaced a use of typename
with class
. I suspect your confusion was because typename
has two meanings -- one as a marker for a kind of template
argument, and another as a disambiguator for a dependent type.
Here the 2nd argument is a pointer, whose type is dependent on the first. The compiler cannot determine if these two conflict without first substituting in the type T
-- and you'll note that they will never actually conflict.
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