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What is printed by the print statements in the program $P1$ assuming call by reference parameter passing?

Program P1()
{
x = 10;
y = 3;
func1(y,x,x);
print x;
print y;
}

func1(x,y,z)
{
y = y + 4;
z = x + y + z
}
1. $\text{10, 3}$
2. $\text{31, 3}$
3. $\text{27, 7}$
4. None of the above
recategorized | 2.5k views
+1
Why it is out of syllabus  taged ?

Here, variable $x$ of func1 points to address of variable $y.$

and variables $y$ and $z$ of func1 points to address of variable $x.$

Therefore, $y=y+4 \implies y=10+4 = 14$

and $z=x+y+z \implies z=14+14+3=31$

$z$ will be stored back in $x.$  Hence, $x=31$ and $y$ will remain as it is. $(y=3)$

Answer is $31, 3$
selected by
+2

should it look like ??

Program P1()
{
x = 10;
y = 3;
func1(&y,&x,&x)
print x;
print y;
}

func1(*x,*y,*z)
{
* y = *y + 4;
*z = *x +* y +* z
}

AS IT IS CALL BY REFERRENCE  ??

+5
^Yes, the result should be same - but change to func1(&y, &x, &x);

But in theory you are using call by value- and the value being a pointer. Using '*' operator this is making the effect of call-by-reference.
+5
@Arjun z = x + y + z ⇒ z = 3 + 14 + 14 = 31  ??? is this right order
0
Hey, actually I can't understand which line is shows that variable x points the address of variable y???
0

In Program P1() where function has been called, the order is func1(y,x,x)}but in function definition it's func1(x,y,z). The order of x and y has been changed keeping the parameter names similar. Since parameter names could be anything inside function definition, x is pointing to address of y, y and z both are pointing to address of x.

0
in the line z=x+y+z why are we not taking x as 3, i mean to say 3+14+10=27?
0
Yes sir why not z=27 is taking
0
Since it is call by reference, so y and z both will point to x. The "y=y+4" statement changes the value of y to 14 and and hence now y and z both will point to x having value 14. Therefore we have to take y and z value as 14 in func1(x,y,z).

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