I have a class as follows,
template<size_t N>
class A
{
typedef int(*ARRAY)[N];
array getArray();
ARRAY array;
};
Then how can I use the type ARRAY in the definition? I have tried this
template<size_t N>
ARRAY A<N>::getArray()
{
return array;
}
But, it tells me that ARRAY is not defined.
ARRAY is depended name, therefor you need to use the typename keyword before it, outside the class scope.
template<size_t N>
typename A<N>::ARRAY A<N>::getArray()
//^^^^^^^^^^^^^
{
return array;
}
or using trailing return
template<size_t N>
auto A<N>::getArray() -> ARRAY
// ^^^^^^^^^^^^^
{
return array;
}
In addition, you have a typo in your class member declaration. The
array getArray();
should be
ARRAY getArray();
However, is there any reason that you do not want to use the std::array?
The problems is that you haven't defined a type ARRAY in the namespace, but rather within the scope of the class template. As such, the unqualified lookup doesn't find a type by that name.
Besides using a qualified name to refer to this member type alias as shown in Bill Lynch's answer, another way is to use a trailing return type in which case the name is looked up within the context of the class:
template<size_t N>
auto A<N>::getArray() -> ARRAY {
return array;
}
P.S. I recommend highly against obfuscating the pointerness (or referenceness) of types by defining type aliases unless the alias is used in generic programming where it might be replaced by a fancy pointer (or proxy reference). In such cases the alias should conventionally be named to signify that it is a "pointer" (or "reference") such as for example in the case of type aliases of iterators and allocators and their type traits in the standard library.
template<size_t N>
typename A<N>::ARRAY A<N>::getArray(){
return array;
}