Is there a difference between the following approaches?
// approach 1
namespace std
{
template<>
void swap<Foo>(Foo& x, Foo& y) // note the <Foo>
{
x.swap(y);
}
}
// approach 2
namespace std
{
template<>
void swap(Foo& x, Foo& y)
{
x.swap(y);
}
}
I stumpled upon this when I tried to specialize swap for my own string type and noticed that swap<::string>
doesn't work, but for a completely different reason :)
A function template starts with the keyword template followed by template parameter(s) inside <> which is followed by the function definition. In the above code, T is a template argument that accepts different data types ( int , float , etc.), and typename is a keyword.
The act of creating a new definition of a function, class, or member of a class from a template declaration and one or more template arguments is called template instantiation. The definition created from a template instantiation is called a specialization.
To instantiate a template function explicitly, follow the template keyword by a declaration (not definition) for the function, with the function identifier followed by the template arguments. template float twice<float>( float original ); Template arguments may be omitted when the compiler can infer them.
- We can define a template for a function that can help us create multiple versions for different data types. - A function template is similar to a class template and it syntax is as follows : template <class T> Return-type functionName (arguments of type T)
Yes, there is. But not in that particular example. If the parameter is not deduced, it can make a difference
template<typename T> void f(typename T::type t);
You cannot specialize that without <type>
because it cannot deduce what T
is from the parameter list.
struct MyType { typedef int type; };
// needs <MyType>
template<> void f<MyType>(int t) { }
Of course in your case, it's the digraph <:
that's meaning the same as [
causing your problem. Put a space like <
::string>
to avoid the problem.
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