Question

Question 3

A. Show the output of the following code.

#include <iostream>

using namespace std;

void magic (int &a, int b, int& c)

{

  a *= 2;

   b = b+2;

   c = c-2;

}

int main ()

{

int x=1, y=3, z=7;

   magic (x, y, z);

cout << "x=" << x << ", y=" << y << ", z=" << z;

   magic (z, y, x);

cout << "x=" << x << ", y=" << y << ", z=" << z;

   return 0;

}

What is the parameter passing scheme for a, b and c in magic( ) function ?

B. Show the output of the following code.

#include <iostream>

using namespace std;

void printarray (int arg[], int length) {

   for (int n=length-1; n>=0; n--)

cout << arg[n] << " ";

   cout << endl;

}

int main ()

{

int firstarray[] = {3, 15, 10};

   int secondarray[] = {2, 4, 6, 8, 10};

   printarray (firstarray,2);

   printarray (secondarray,3);

   return 0;

}

Answer 1

Question 3 A):

#include <iostream>
using namespace std;
void magic (int &a, int b, int& c)
{
a *= 2;
b = b+2;
c = c-2;
}
int main ()
{
   int x=1, y=3, z=7;
magic (x, y, z);
cout << "x=" << x << ", y=" << y << ", z=" << z;
magic (z, y, x);
cout << "x=" << x << ", y=" << y << ", z=" << z;
return 0;
}

Screesnhots:

Output:

The parameter passing mechanism is both call by value and call by reference too.

Since in magic() method x reference(address) and z reference(addess) is used, the function magic() for x and z is call by reference method.

Since in magic() method y is used directly (formal parameter) , the function magic() for y is call by value method.

Parameter scheme for x is call by reference.

Parameter scheme for y is call by value.

Parameter scheme for z is call by reference.

Quesiton 3 B):

#include <iostream>
using namespace std;
void printarray (int arg[], int length)
{
for (int n=length-1; n>=0; n--)
cout << arg[n] << " ";
cout << endl;
}
int main ()
{
   int firstarray[] = {3, 15, 10};
int secondarray[] = {2, 4, 6, 8, 10};
printarray (firstarray,2);
printarray (secondarray,3);
return 0;
}

Screeshots:

Output: