Template in C++ with raw type

Question

In Java, we can do something like this:

ArrayList arrayList = new ArrayList<String>(); 
// Notice that String is not mentioned in the first declaration of array

AS OPPOSED TO

ArrayList<String> arrayList = new ArrayList<String>(); 

How can we something in similar in C++?

Answer 1

Not in exactly the way you've written.

What you can do is one of the following, depending on what you're actually trying to accomplish:

• In C++11, you can use auto to automatically adapt to the type: auto = new ArrayList<String>(); . This doesn't give you polymorphism, but it does save you typing the typename on the left hand side.
• If you want polymorphism, you can add a level to your class hierarchy, and make the left-hand side point to a parent class.

Here's an example of the second approach:

class IArrayList   // define a pure virtual ArrayList interface
{
     // put your interface pure virtual method declarations here
};

template <typename T>
class ArrayList : public IArrayList
{
     // put your concrete implementation here
};

Then, you could say in your code:

IArrayList* arrayList1 = new ArrayList<string>();
IArrayList* arrayList2 = new ArrayList<double>();

...and so on.

Answer 2

在c ++中，不能使用vector array = new vector<string>() ，但是在c ++ 11中，可以使用auto关键字： auto p = new vector<string>() ，它与vector<string> *p = new vector<string>()相同vector<string> *p = new vector<string>()希望我的回答可以对您有所帮助。