Question

Trace the following program. What is the output after execution?

typedef double T;

class PlainBox{

private:

T item;

public:

PlainBox();

  PlainBox(const T& theItem);

void setItem(const T& itemItem);

T getItem() const;

friend ostream & operator<<(ostream & out, const PlainBox & theBox);

  bool operator<(const PlainBox & theBox);

};

PlainBox::PlainBox(){}

PlainBox::PlainBox( const ItemType & theItem){

item = theItem;

}

void PlainBox:: setItem(const T& theItem){

item = theItem;

}

T PlainBox:: getItem () const

{

return item;

}

//remaining implementation is not shown

source.cpp

#include "plainbox.h"

...

void main(){

PlainBox box1, box2(7.9);

    box1.setItem(1.5);

cout<< box1.getItem()<<endl;

  cout<< box2.getItem()<<endl;

if( box1.getItem() < box2.getItem() )

   box2.setItem( box1.getItem() );

  cout<< box1.getItem()<<endl;

  cout<< box2.getItem()<<endl;

double d =  box1.getItem() * 10;

box2.setItem(d);

  cout<< box1.getItem()<<endl;

  cout<< box2.getItem()<<endl;

}

Answer 1

Here is the solution to above program . Please read step by step for full explanation also give a thumbs up if you like the solution

1) The above class is of template type hence it can make plain Box containg any datatype of members

The first output statement in main are

PlainBox box1, box2(7.9);

box1.setItem(1.5);

cout<< box1.getItem()<<endl;

cout<< box2.getItem()<<endl;

Here we are using constructor and mutator/accessor method to set box1 item to 1.5 and box2 item to 7.9 hence output will be

1.5

7.9

2)

if( box1.getItem() < box2.getItem() )

   box2.setItem( box1.getItem() );

  cout<< box1.getItem()<<endl;

  cout<< box2.getItem()<<endl;

In the above lines we are comparing if box1 is lesser than box2 which is true since 1.5<7.9 therefore box 2 item will be set to box1 item value hence both box1 and box2 hold 1.5 as value

there output will be

1.5

1.5

3)

double d =  box1.getItem() * 10;

box2.setItem(d);

cout<< box1.getItem()<<endl;

cout<< box2.getItem()<<endl;

In the final statements we are again modifying the value of box2 by box1*10 hence the value of box2 will be 1.5*10 =15 and box1 value will be 1.5 so output will be

1.5

15

HENCE COMBINING STEP 1 , 2 and 3 we get final output as

1.5

7.9

1.5

1.5

1.5

15