Question

p3_unzip :: [(a,b)] -> ([a], [b]) Write a function that takes a list of two-tuples and returns a tuple of two lists -- the first containing the first element of each tuple, and the second the second element (i.e., the reverse of "zip"). Examples: > p3_unzip [('a',1),('b',5),('c',8)] ("abc",[1,5,8])

Answer 1

Answer:

C++.

Explanation:

#include <iostream>

#include<tuple>

using namespace std;

////////////////////////////////////////////////////////////

tuple<char*, int*> p3_unzip(tuple<char, int>* list, int count) {

   char* list1 = new char[count];

   int* list2 = new int[count];

   

   for (int i=0; i<count; i++) {

       tuple<char, int> temp = list[i];

       list1[i] = get<0>(temp);

       list2[i] = get<1>(temp);

   }

   

   tuple<char*, int*> new_tuple;

   get<0>(new_tuple) = list1;

   get<1>(new_tuple) = list2;

   

   return new_tuple;

}

int main() {

   tuple<char, int>* list;

   int count = 0;

   ////////////////////////////////////////

   cout<<"How many tuples? ";

   cin>>count;

   list = new tuple<char, int>[count];

   cout<<endl;

   

   char a;

   int b;

   tuple<char, int> temp;

   for (int i=0; i<count; i++) {

       cout<<"tuple "<<i+1<<" char: ";

       cin>>a;

       get<0>(temp) = a;

       

       cout<<"tuple "<<i+1<<" int: ";

       cin>>b;

       get<1>(temp) = b;

       

       list[i] = temp;

       cout<<endl;

   }

   ////////////////////////////////////////

   tuple<char*, int*> new_tuple = p3_unzip(list, count);

   char* list1 =  get<0>(new_tuple);

   int* list2 = get<1>(new_tuple);

   

   cout<<"Returned tuple:([";

   for (int i=0; i<count-1; i++) {

       cout<<list1[i]<<", ";

   }

   

   cout<<list1[count-1]<<"],[";

   

   for (int i=0; i<count-1; i++) {

       cout<<list2[i]<<", ";

   }

   

   cout<<list2[count-1]<<"])";

   ////////////////////////////////////////

   delete[] list, list1, list2;

   return 0;

}