dictionary - Looking for proper container / functions for enumerable in c++ -

i trying convert code c# c++ lack of dictionary tables/enumerables etc making me difficult result needed in c++. can type of container/methods use in c++ result needed?

thanks in advance.

find c1 , it's count group c1 c2 > 0 , c3 < 4 order c1

table(c1,c2,c3)  ( number of rows expected not finite - - can't use array structure ) 5 1 2 4 2 3  --> edited line make list 4 4 3 4 0 1  --> ignore row c2=0 3 1 3   2 1 5  --> ignore row c3 > 4 ..... 

.....

expected output(number of rows meeting criteria each c1): 3 1 4 2 5 1 

you need @ least:

• a struct hold each c1/c2/c3 tuple (or std::tuple if use c++11).
• a std::vector (array-like container, dynamic size) hold tuples.
• a std::map (sorted associative container) act dictionary compute output.

i believe enough started, if have specific problem when writing code don't hesitate ask new questions.

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edit according comments:

you're not missing much, elvena's solution almost need except lacks vector container store objects. quite straightforward:

#include <iostream> #include <map> #include <vector> #include <tuple>  int main() {     std::vector<std::tuple<int, int, int>> values;     while (you_have_more_data) {         int c1, c2, c3;         // somehow read c1, c2, c3 cin/file/whatever         values.push_back(std::make_tuple(c1, c2, c3));     }      std::map<int, int> dict;     // iterate on vector     (auto = values.begin(); != values.end(); ++i) {         // *i (dereferencing iterator) yields std::tuple<int, int, int>         // use std::get access individual values in tuple         // 0 => c1; 1 => c2; 2 => c3 (same order in std::make_tuple)         if (std::get<1>(*i) > 0 && std::get<2>(*i) < 4)             dict[std::get<0>(*i)] += 1; // see std::map::operator[]     }      // iterate on map , print items     (auto = dict.begin(); != dict.end(); ++i)         // *i (dereferencing iterator) yields std::pair<int, int>         // writing (*i).first cumbersome         // let's write i->first instead (this same, different notation)         std::cout << i->first << " " << i->second << std::endl;      return 0; }