Question

C++ ONLY! Consider the following code and answer the questions in the table below.

#include <iostream>

template <typename T, int N=10>

class Array {

private:

  T array[N+1];

public:

  Array() {

    for(int i=0; i<N+1; i++) array[i] = 0;

  }

  

  T& operator [] (int index) {

    if (index>N || index < 0)

      return array[N];

    else

      return array[index];

  }

  template <typename S>

  Array<T,N>& operator= (S &other) {

    for(int i=0; i<N+1; i++)

      array[i] = other[i];    

    return *this;

  }

};

template <typename T, typename S> T operator+ (T &t, S &s)

{

  T res;

  for(int i=0; i<10; i++)

    res[i] = t[i] + s[i];  

  return res;

}

template <typename T, int N> std::ostream& operator << (std::ostream& out, Array<T,N> &array)

{

  out << "{ ";

  for(int i=0; i<N; i++)

    {

      if (i>0) out << ", ";

      out << array[i];

    }

  out << "}\n";

  return out;

}

int main() {

   // code snippet runs here

   return 0;

}

Prompt Snippet Output / Error / Explanation What is the template specialization for variable x? Array int> x; 2 What is the output? Array<int, 5> x; std::cout << x; 3 What is the output? Array<int i; Array<double> d; i[1] = 10.5; d[1] = auto res = d+i; std::cout << res; 5.5; 4 What type is res deduced to be? Array int> i; (Hint: res must match the type on Array<double> d; the right side of the assignment.) i[1] = 10.5; d[1] = 5.5; auto res = d+i; std::cout << res; 5 What is the output? Array int> cvt; char hello [10] = "Hello!"; cvt = hello; std::cout << cyt; 6 What is the specialization of operator= that is deduced by the compiler in this code snippet? Array int> cvt; char hellol) = "yoyoyoyo!"; cvt = hello; 7 What is the output? Array<int i; for (int j=-1; j<100; j++) ilj] = j; std::cout << i; 8 What is the output? Briefly explain how the code is calculating that result. Array<int i; for (int j=-1; j<12; j++) ilj] = j;

Answer 1

Question-1

This code snippet willnot throw any error nor its shows any ouput.

Given Array<int> x;

Here the template specialization for variable is int data type.
T in the template definition is substituted with int.

Question-2

Array<int,5> x;
std::cout<<x;

Output is

{ 0, 0, 0, 0, 0}

Question-3

Array<int> i;
Array<double> d;
i[1]=10.5; d[1] = 5.5;
auto res = d+i;
std::cout<<res;

Output is

{ 0, 15.5, 0, 0, 0, 0, 0, 0, 0, 0}

Out put is 15.5 because though we give i[1] =10.5 it will take it as 10, because it is int type. This will be added to 5.5 and gives 15.5

Question-4

Array<int> i;
Array<double> d;
i[1]=10.5; d[1] = 5.5;
auto res = d+i;
std::cout<<res;

res must be deduced to Array<double> type since the result is of double type.

Then the code snippet is

Array<int> i;
Array<double> d;
i[1]=10.5; d[1] = 5.5;
Array<double> res = d+i;
std::cout<<res;

Question-5

Array<int> cvt;
char hello[10] = "yoyoyoyo!";
cvt= hello;

This code snippet willnot throw any error nor its shows any ouput.

Here cvt is an int Array, but hello is assigned to it.
So compiler converts the ASCI value of each character and assigns it to particular blocks of the array.

Note:- Please ask remaining questions as another question.

Feel free to ask any doubts, if you face any difficulty in understanding.

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