Question

2:31 $32% X 324 325 14) Given the following function definition: void calc (int a, int& b) intc; C = a + 2; a = a* 3; b = c + a; What is the output of the following code fragment that invokes calc? int x = 1; int y = 2; int z = 3; calc(x,y); cout << x <<""<< y «""<< Z << endl; A) 123 B) 163 C) 3 63 D) 1149 E) None of these

Can someone please explain to me this? I'm having a hard time understanding it. thank you

Answer 1

Output:

E) None of these

The actual output is 1 5 3 (which is not mentioned in the given options )

" As you pass any argument by using call by reference then changes made to the parameter affect the passed argument"

Here, x=1 and y =2 so as you call the calc() where a is normal variable and b is a reference variable so the value of y is assigned to b ie value of b is now 2 so inside the function,

c= a* b = 1 * 2 = 2

a = a * 3 = 1 * 3 = 3

b = c + a = 2 + 3 = 5

As I said earlier

" As you pass any argument by using call by reference then changes made to the parameter affect the passed argument"

so the value of b is 5 so the value of y also becomes 5

so as you print the values of x, y and z then the output is x=1, y=5, z=3.

If you remove & ie if treat b as a normal variable then the output will be 1 2 3,

because calc() function doesn't return anything so it will display the values of x,y, and z;