Given the following function definition:
what is the output of the following code fragment that invokes calc?
1 2 3 1 6 3 3 6 3 1 14 9
none of these

The question is incomplete! Complete question along with its step by step answer is provided below!

Question:

Given the following function definition:

void calc (int a, int& b)

{

int c;

c = a + 2;

a = a * 3;

b = c + a;

}

x = 1;

y = 2;

z = 3;

calc(x, y);

cout << x << " " << y << " " << z << endl;

What is the output of the following code fragment that invokes calc?

a. 1 2 3

b. 1 6 3

c. 3 6 3

d. 1 14 9

e. None of these

b. 1 6 3

Explanation:

In the given problem we have a function void calc which takes two input arguments a and b and updates its values according to following equations

c = a + 2;

a = a * 3;

b = c + a;

Then we call this function calc(x,y) by providing test values of

int x = 1;

int y = 2;

int z = 3;

and the output returns the values of x, y and z

cout << x << " " << y << " " << z << endl;

Lets find out what is happening here!

When the program runs we provide x=a=1 and y=b=2

c=a+2=1+2=3

a=a*3=1*3=3

b=c+a=3+3=6

So the updated values of a=x=3 and b=y=6?

NO!

The updated values are a=x=1 and b=y=6

WHY?

There are two ways to pass values

1. Pass by values -> value cannot change  (int a)

2. Pass by reference -> value can change (int& b)

Look at the function void calc (int a, int& b) ;

Here we are passing (int a) as a value and (int& b) as a reference, therefore x remains same x=1 and y gets changed to updated value y=6 and z remains same as z=3 since it wasnt used by function calc(x,y)

The right answer is:

b. 1 6 3

x=1, y=6, z=3

answer; i believe that the correct answer is (calculated field);

add the following answer