Inheritance instead of typedef

Question

C++ is unable to make a template out of a typedef or typedef a templated class. I know if I inherit and make my class a template, it will work.

Examples:

// Illegal
template <class T>
typedef MyVectorType vector<T>;

//Valid, but advantageous?
template <class T>
class MyVectorType : public vector<T> { };

Is doing this advantageous so that I can "fake" a typedef or are there better ways to do this?

similar question, same answers: http://stackoverflow.com/questions/293988/generic-vector-of-vectors-in-c — Johannes Schaub - litb, Jan 14 '09 at 04:08
Your second example is invalid too. You will have to write your own constructors for MyVectorType since you can not inherit them. — Sebastian, Jul 06 '11 at 08:59

score 14 · Accepted Answer · answered Jan 14 '09 at 03:07

14

C++0x will add template typedefs using the using keyword.

Your solution declares a new type, not a type "alias", e.g. you cannot initialize a MyVectorType & (reference) with a vector<T>. This might not be a problem for you, but if it is, but you don't want to reference vector in your code, you can do:

template <typename T>
class MyVectorType {
public:
  typedef std::vector<T> type;
};

answered Jan 14 '09 at 03:07

Magnus Hiie

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std::vector has a non-virtual destructor so you need to be careful not delete a MyVectorType via a std::vector* – jk. Dec 23 '09 at 09:25
I was not proposing to use MyVectorType as the actual datatype, but rather MyVectorType::type, which is a typedef for std::vector, so no danger here. – Magnus Hiie Feb 04 '10 at 23:29

score 10 · Answer 2 · answered Jan 14 '09 at 03:06

10

Inheritance is not what you want. A more idiomatic approach to emulating a templated typedef is this:

template <typename T> struct MyVectorType {
    typedef std::vector<T> t;
};

Refer to the typedef like this:

MyVectorType<T>::t;

answered Jan 14 '09 at 03:06

Dave Ray

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Max Lybbert · Answer 3 · 2009-12-23T09:19:49.417

1

C++ is unable to make a template out of a typedef or typedef a templated class.

That depends on what you mean by typedef: std::vector<size_t> is legal -- even though size_t is a typedef -- and std::string is a typedef for std::basic_string<char>.

edited Dec 23 '09 at 09:19

answered Jan 14 '09 at 08:20

Max Lybbert

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score 1 · Answer 4 · answered Jul 31 '18 at 16:53

What you should use depends on how it should behave (overload resolution for example):

If you want it to be a new distinct type
```
class Foo : public Whatever {};
```
However, you can now pass pointers or references of Foo, to functions accepting Whatever, so this becomes a hybrid. So in this case, use private inheritance, and using Whatever::Foo or custom implementations, to expose the methods you want.
If you want a type synonym
```
using Foo = Whatever;
```

The former has the advantage that it is possible to predeclare a class, but not a typedef.

Inheritance instead of typedef

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