Question

In C++

void func( int& num1, int* num2)

{

              int temp = *num2;

*num2 = num1;

num1 = temp - *num2;

}

Given the function definition above what would be the result of the following code?

int val1 = 5;

int val2 = 8;

func(val1, &val2);

cout << val1 << “,” << val2 << endl;

Answer

a) 13, 8   

b) 8, 13

c) An error message

d) 3, 5

e) 8, 5

Answer 1

void func( int& num1, int* num2)

{

              int temp = *num2;

*num2 = num1;

num1 = temp - *num2;

}

Given the function definition above what would be the result of the following code?

int val1 = 5;

int val2 = 8;

func(val1, &val2);

cout << val1 << “,” << val2 << endl;

Answer: d. 3,5

Explanation:

//here func takes the address of val1 and the reference to the address of val2 as parameters
void func( int& num1, int* num2)
{
   //now &num1 will have the address of val1 and num2 will have the address of val2
  
int temp = *num2;//temp will point to the value in num2 i.e., 8
   *num2 = num1; //num2 will point to the value in num1 i.e., 5
   num1 = temp - *num2; //now num1 8 - 5 i.e, 3
  
   //finally num1 will have the value 3 and num2 will have the value 5
}

int main()
{
   int val1 = 5;
   int val2 = 8;
   func(val1, &val2); //pass val1 and address of val2 to func
   cout << val1 << ","<< val2 << endl; //prints 3 and 5
   return 0;
}

Output: