Getting the type specifier given an object

Question

I am trying to write a template function that will take an STL container and will display all the occurrences of the elements in it and the number that they have occurred. I am planning to use a map, iterate through the container and either add a new element if it does not exist or increment the count of occurrences for the element.

Declaration:

template < typename Container_t >
void findOccurrences (const Container_t& inContainer);

My question is: can I somehow get the type specifier of the element that the container holds? So when I create my map the key value would be the element in the inContainer. Something like :

map < typeid ( * inContainer.begin()), int > occurrences;

Or would I have to change my template to something like this:

template < typename Container_t , typename Element_t >
void findOccurrences ( const Container_t & inContainer , Element_t dummy )
{
  map < Element_t , int > occurrences;
}

Thanks

score 4 · Accepted Answer · answered Jun 26 '11 at 19:56

4

How about something like this:

#include <map>
#include <iterator>

template <typename Iter>
void histogram(Iter begin, Iter end)
{
  typedef typename std::iterator_traits<Iter>::value_type T;

  std::map<T, size_t> h;

  while (begin != end) ++h[*begin++];

  // now h holds the count of each distinct element
}

Usage:

std::vector<std::string> v = get_strings();
histogram(v.begin(), v.end());

answered Jun 26 '11 at 19:56

Kerrek SB

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`iterator_traits` also requires C++0x, does it not? – Ben Voigt Jun 26 '11 at 20:14
@Ben: Err... Nope, iterator_traits aren't new to C++0x – Armen Tsirunyan Jun 26 '11 at 20:17
@Ben: No, `iterator_traits` has always been there – Xeo Jun 26 '11 at 20:17
@Space_C0wb0y: Yay for genericity! Should I have said `const Iter end`, though? @Ben: You can't use iterators without [iterator traits](http://www.cplusplus.com/reference/std/iterator/iterator_traits/)! – Kerrek SB Jun 26 '11 at 20:24

score 3 · Answer 2 · answered Jun 26 '11 at 19:51

3

You want typename Container_t::element_type

That is,

std::map <typename Container_t::element_type, int>

answered Jun 26 '11 at 19:51

Armen Tsirunyan

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Names ending in `_t` are reserved. – Ben Voigt Jun 26 '11 at 19:53
@Ben: I don't know if that's entirely true in any scope... but then it's not that relevant to this question. Thanks for the info anyway :) – Armen Tsirunyan Jun 26 '11 at 19:54
1

@Ben: I've never seen anything in the Standard about reserving those names. – Puppy Jun 26 '11 at 20:06
http://stackoverflow.com/questions/1186072/naming-scheme-for-typedefs/1186143#1186143 – Ben Voigt Jun 26 '11 at 20:20
@Ben: It's about global scope, ain't it? In this situation it is not a problem at all – Armen Tsirunyan Jun 26 '11 at 20:22
@Armen: That's probably true, although a future library update could potentially define a macro by that name, breaking things. – Ben Voigt Jun 26 '11 at 21:16
@Ben: I don't think that a name ending with _t is a suitable name for a macro ) – Armen Tsirunyan Jun 26 '11 at 21:17

Ben Voigt · Answer 3 · 2011-06-26T20:12:00.260

1

With C++0x, it's really easy:

map<decltype(*c.begin()), int> occurrences;

With C++03, you probably need to use a typedef from the container:

template<typename Container>
// ...
map<Container::element_type, int> occurrences;

edited Jun 26 '11 at 20:12

answered Jun 26 '11 at 19:52

Ben Voigt

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It's `decltype`, not `declspec`. – Puppy Jun 26 '11 at 20:06
@DeadMG: Right you are! Thanks. – Ben Voigt Jun 26 '11 at 20:12

score 0 · Answer 4 · answered Jun 26 '11 at 19:51

0

Please take a look at "RTTI" (run-time type information)

http://en.wikipedia.org/wiki/Run-time_type_information

I hope this helps.

answered Jun 26 '11 at 19:51

feathj

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Getting the type specifier given an object

4 Answers4