Template functions with STL Vectors break with new allocator

Question

I have a ton of functions that are defined as something like:

template<typename T>
void myFunction1(vector<T>& vin);

The point being, I input an STL vector and do some work.

I have recently needed to change the default allocator to a customer allocator. This seems to break every function unless I change the definitions to:

template<typename T, typename Alloc>
void myFunction1(vector<T,Alloc>& vin);

To make it more complicated, I will not be using the overloaded allocator in all cases..

Does this mean I have to rewrite EVERY function with two definitions, one with the template for the allocator and the other definition without the allocator? I really hope this isn't the answer...

juanchopanza · Answer 1 · 2012-11-15T15:41:43.147

2

A simpler solution could be to implement your functions in terms of iterators, as is done in standard library algorithms:

template <typename Iterator>
void myFunction1(Iterator1 first, Iterator2 last) { .... }

edited Nov 15 '12 at 15:41

answered Nov 15 '12 at 15:35

juanchopanza

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Kerrek SB · Accepted Answer · 2012-11-15T15:53:02.963

1

It's entirely adequate to have one single function template that respects the full class template. Hypothetically, this will do:

template <typename T, typename Alloc>
void myFunction1(std::vector<T, Alloc> & v);

Every vector has those two arguments, no matter whether the allocator one is defaulted or not.

However, a more fruitful idiom is actually to make the entire container a template:

template <typename V>
void myFunction1(V & v)
{
    typedef typename V::value_type value_type;
    // ...
}

edited Nov 15 '12 at 15:53

answered Nov 15 '12 at 15:36

Kerrek SB

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You shouldn't need to go to variadics. The library requirements say that there's a template named `vector` that takes two type arguments; while implementations are allowed to add extra arguments, they aren't allowed to make `vector` not work. The place where you can run into trouble is if you try to write your own declaration of a standard library type. Using a standard library type is simply required to work as documented. – Pete Becker Nov 15 '12 at 15:51
@PeteBecker: You're right. I was thinking of a different situation (specialization) where the "allowed extra arguments" come to bite you. Let me remove it from this post, though! – Kerrek SB Nov 15 '12 at 15:52
@PeteBecker: If you're curious, that's currently an open issue in the [pretty printer](http://stackoverflow.com/questions/4850473/pretty-print-c-stl-containers): I want to provide specialized delimiters for sets, but I don't have a portable solution. – Kerrek SB Nov 15 '12 at 16:28

score 0 · Answer 3 · answered Nov 15 '12 at 15:33

0

I think you have too change your template. But the good news is you can change your template to this:

template <typename MyVector> 
void myFunction1(MyVector& vin);

answered Nov 15 '12 at 15:33

0x26res

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Of course, that change means that the function can be called with any type at all, not just a `vector`. – Pete Becker Nov 15 '12 at 15:46

score 0 · Answer 4 · answered Nov 15 '12 at 15:38

0

You don't have to provide two overloads for each function template. Just provide the two-parameter one. std::vector has two template parameters, so both will be deduced just fine.

answered Nov 15 '12 at 15:38

Angew is no longer proud of SO

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Template functions with STL Vectors break with new allocator

4 Answers4