Make template class method parameter type same as class template parameter (including void)
I would like to make a class method have a signature that matches the class template parameter, e.g., for a Class<T>
, there’s Class<T>::Method(T t)
if T
is not void
, and Class<T>::Method()
if T
is void
.
Make template class method parameter type same as class template parameter (including void)
I would like to make a class method have a signature that matches the class template parameter, e.g., for a Class<T>
, there’s Class<T>::Method(T t)
if T
is not void
, and Class<T>::Method()
if T
is void
.
Make template class method parameter type same as class template parameter (including void)
I would like to make a class method have a signature that matches the class template parameter, e.g., for a Class<T>
, there’s Class<T>::Method(T t)
if T
is not void
, and Class<T>::Method()
if T
is void
.
Make template class method parameter type same as class template parameter (including void)
I would like to make a class method have a signature that matches the class template parameter, e.g., for a Class<T>
, there’s Class<T>::Method(T t)
if T
is not void
, and Class<T>::Method()
if T
is void
.
Make template class method parameter type same as class template parameter (including void)
I would like to make a class method have a signature that matches the class template parameter, e.g., for a Class<T>
, there’s Class<T>::Method(T t)
if T
is not void
, and Class<T>::Method()
if T
is void
.
C++ enable_if class specialization not invoked
I have the following example on trying to use enable_if for instantiating 2 different specializations of the class A, but I am not understanding it correctly as they are not getting invoked, any suggestions ? thanks in advance.