I just learned compile-time array declaration via template, with the non-type parameter. so in this approach, we pass constant and array with size gets declared. like
#include<iostream>
using namespace std;
template<int size_of_array>
class arr
{
public:
int arr[size_of_array];
int size_of() { return size_of_array; }
};
int main()
{
arr<4> arr1;
cout << arr1.size_of();
return 0;
}
Can we do something like variable value assigning at run time like if we remove arr[] from there. like
#include<iostream>
using namespace std;
template<int size_of_array>
class arr
{
public:
int size_of_array;
int size_of() { return size_of_array; }
};
int main()
{
arr<4> arr1;
cout << arr1.size_of();
return 0;
}
Can we do that? or why not? Then maybe there is some thing about an array and a variable declaration which i don't know. Thank you in advance.
Absolutely! You can expose the template parameters by writing a function to pass them through.
Your attempt only failed because:
There's really no need for an additional data member here anyway, as the template argument is accessible throughout the class definition.
So, for a non-type template parameter like yours, just:
template <int size_of_array>
class arr
{
public:
static int size_of() { return size_of_array; }
};
Now arr<42>::size_of() is 42!
(I've made it static, not because you need to, but because in this example it makes sense; you could alternatively make it a non-static but const member function.)
And for a type:
template <typename T>
class arr
{
public:
using array_size_t = T;
};
Now arr<T>::array_size_t is the type T!
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