Consider the following C++ template class.
#include
using namespace std;
template
class SortedList
{
public:
SortedList()
{size = 0;}
void insert(T item);
friend ostream& operator<<(ostream& out, const SortedList& list)
{return list.put(out);}
private:
ostream& put(ostream& out) const;
T list[length];
int size;
};
template
void SortedList::insert(T item)
{
if (size == length)
throw exception("List Full");
int i = size - 1;
while (i >= 0 && item < list[i])
{
list[i+1] = list[i];
i--;
}
list[i+1] = item;
size++;
}
template
ostream& SortedList::put(ostream& out) const
{
for (int i = 0; i < size; i++)
cout << list[i] << " " ;
cout << endl;
return out;
}
int main()
{
int values[] = {5, 1, 7, 8, 11, 2};
SortedList list;
for (int i = 0; i < 6; i++)
list.insert(values[i]);
cout << list;
return 0;
}
The class SortedList cannot be instantiated for any arbitrary type. For example, consider the following instantiation for a wrapper integer class.
class Int
{
public:
Int(int i) {this->i = i;}
private:
int i;
};
int main()
{
Int values[] = {Int(5), Int(1), Int(7), Int(8), Int(11), Int(2)};
SortedList list;
for (int i = 0; i < 6; i++)
list.insert(values[i]);
cout << list;
return 0;
}
Explain why the second implementation fails. What must be added to that class so this program will compile? Suppose this program were written in Java. Explain how Java allows the constraints on a generic type parameter to be specified and how they would be specified in this case
Java does have one limitation, however. Although wrapper classes can be used to instantiate generic type parameters, primitive types cannot. Explain why.