Check int or list

Question

Can anybody please let me know the best way to accomplish this.

Say, I have a template function like

template<typename ARGUMENT>
void get_result(ARGUMENT &ag)
{

// arg can be a single object of a particular object or list of objects of that particular class. 
//rest

}

Is there a way that I can check if the &ag is a single object or list of objects. Also, with the given template interface.

It does not matter if the answer is by template specification in some way by a class interface. Only thing is I do not want to specify the object type or list type.

Ex. ag = int or ag = list

CB

Actually any container. Also it would be good if I can select the return type accordingly. If one object single object return and if argument is a container container return type. On the other hand inside the function it is the same routine for objects other than iterating when it is a container. It would be better if I can have the shortest code. — cow boy, Jul 29 '13 at 09:48
Conceptually, either the arguement to your function is a container or it's not. If you want to mix the two within the same function, it sounds like bad design. Is your intention to treat passing a single element, as effectively a container of size 1? — Neil Kirk, Jul 29 '13 at 11:05
@NeilKirk Why is it a bad design. This is just to provide the user with the same interface to give arguements as a single object (say geometric point) and a set of objects (set of points). — cow boy, Jul 29 '13 at 11:22
Provide two functions, one that takes a single object and one that takes a group. — Neil Kirk, Jul 29 '13 at 11:29
The library is template and I have to respect the design and can not provide two functions. — cow boy, Jul 29 '13 at 11:40

jrok · Answer 1 · 2013-07-29T11:11:13.060

7

Hmm, maybe all you need is simple overloading?

template<typename ARGUMENT>
void get_result(ARGUMENT& ag);

template<typename ARGUMENT>
void get_result(std::list<ARGUMENT>& ag);

Edit:

Reading your comments, I have a feeling you're trying to overdesign your function and give it to many responsibilities.

I think you'd be best off with the first overload only. Whenever you need to apply the function to a whole range, use for_each.

edited Jul 29 '13 at 11:11

answered Jul 29 '13 at 09:16

jrok

54,456
9
109
141

This might well work. However I would like to to have something like, ARGUMENT::iterator itr = ag.begin() after checking if the argument is from a object container and apply the same routine as for single objects. So, if I can have it in one function that will be great. – cow boy Jul 29 '13 at 10:07
@cowboy: You can achieve that by treating a non-container as a container with one element. But that might actually be more work. – MSalters Jul 29 '13 at 10:55

score 2 · Accepted Answer · edited May 23 '17 at 11:49

You could disambiguate with some type traits:

#include <type_traits>

template<typename T>
T get_result(T arg) 
{ 
    return detail::get_result(arg, typename std::is_arithmetic<T>::type() ); 
}

namespace detail {
    template<typename T>
    T get_result(T arg, std::false_type /* dummy */) { }

    template<typename T>
    T get_result(T arg, std::true_type /* dummy */) {}
}

See here
This trait clearly just pulls out the numeric types, rather than a container. The return type will take some work. There are ideas for detecting a container type in the answers here and here

P0W · Answer 3 · 2013-07-29T09:28:20.717

1

Do you mean this ?

template<typename ARGUMENT>
void get_result(ARGUMENT &ag)
{
   std::cout<<typeid(ARGUMENT).name();  //Check the value 

/*This returns a name of the type in question.
 It may satisfy your criteria.*/
}
//Header :#include <typeinfo>

So,

int x=12;
list <int> l;
get_result<list<int>>(l); // Outputs: St4listIiSaIiEE

get_result<int>(x); //Outputs: i

edited Jul 29 '13 at 09:28

answered Jul 29 '13 at 09:21

P0W

46,614
9
72
119

Thnx. I would rather like a solution without typechecking. – cow boy Jul 29 '13 at 10:08

score 1 · Answer 4 · answered Jul 29 '13 at 10:52

1

SFINAE:

template<typename T>
T get_result(T arg, typename T::value_type* = 0) 
{ 
    // Container.
}

template<typename T>
T get_result(T arg, ...) 
{ 
    // Not a container. Vararg overloads rank lower in overload resolution.
}

answered Jul 29 '13 at 10:52

MSalters

173,980
10
155
350

But this won't error if nonsense extra arguements are provided? – Neil Kirk Jul 29 '13 at 11:03
@NeilKirk: In that case, wrap it in a `namespace detail` (see doctorlove's answer) – MSalters Jul 29 '13 at 11:04
@MSalters Thnx. will check this out. – cow boy Jul 30 '13 at 12:45

Check int or list

4 Answers4