Test by Bikram | Mock GATE | Test 2 | Question: 14

Question

Consider the following code snippet.

void first(int p, int t)
{
    p +=  t;
    t +=  p;
}

main()
{
    int p = 4;
    first(p , p);
}

What is the final value of $p$ in both call by value and call by reference respectively ?

• A.   $4$ and $12$ 
• B.    $5$ and $12$  
• C.    $12$ and $16$  
• D.    $4$ and $16$

Answer 1

In call by value, function first() operate on copy of p, hence value of p remains unchanged.

In call by reference, first value of p changes to 4+4, then it changes to 8+8, hence 16.

Comments

Can't find x in whole program

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it is p, corrected , thanks :)

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@Bikram veteran

how   solve this give  picture solution

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@devendra

Here p is Integer, therefore p will store 4.

in call by value, p wont change, so p=4.

 

in call by reference, any computation will effect p, here both p and t are referring to same location.
p+=t, this will make value of p =8.
now of course t is also 8, as both referring to same location.

t+=p, this will make final value of t or p as 16. (value of t and p means value refereed by t and p.)

Hence it becomes 4 and 16, option D.