in Operating System edited by
29,940 views
107 votes
107 votes
Consider the following code fragment:
if (fork() == 0)
{
   a = a + 5;
   printf("%d, %p n", a, &a);
}
else
{
   a = a - 5;
   printf ("%d, %p n", a,& a);
}

Let $u,v$ be the values printed by the parent process and $x,y$ be the values printed by the child process. Which one of the following is TRUE?

  1. $u = x + 10  \text{ and } v = y$
  2. $u = x + 10 \text{ and } v != y$
  3. $u + 10 = x \text{ and } v = y$
  4. $u + 10 = x \text{ and } v != y$
in Operating System edited by
29.9k views

4 Comments

 please confirm this – 

before any modification,logical as well as physical address will be same for the variable to be modified.but after modification of the variable,logical address will remain same but physical address will be different due to COW.

like here if we can get to know about physical address of u & v after end of this program then they are not equal,but before starting they were equal.

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2
His name is Chester.

He is god of OS
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8 Answers

74 votes
74 votes
Best answer

It should be Option C.

#include<stdio.h>
#include<stdlib.h>
void main()
{
    int a =100;
    if(fork()==0)
    {
     a=a+5;
     printf("%d %d \n",a,&a );
    }
   else
    {
     a=a-5;
     printf("%d %d \n",a,&a );
    }
}

Output:

Fork returns $0$ when it is a child process.

if ( fork == 0)              

Is true when it is child . Child increment valule of a .

In the above output:

  • $95$ is printed by parent : $\mathbf{u}$
  • $105$ is printed by child  : $\mathbf{x}$
    $\mathbf{\Rightarrow u+10=x}$

The logical addresses remains the same between the parent and child processes.

Hence, answer should be:

 $\mathbf{ u+10=x}$  and $\mathbf{v=y}$

edited by
by

4 Comments

@jatin

your logic seems to be correct for me
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edited by
why not  v!=y???
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Both variables can never be at same location v can never be equal to y.
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33 votes
33 votes
(c) is the answer. Child is incrementing a by 5 and parent is decrementing a by 5. So, x = u + 10.

During fork(), address space of parent is copied for the child. So, any modifications to child variable won't affect the parent variable or vice-verse. But this copy is for physical pages of memory. The logical addresses remains the same between the parent and child processes.

4 Comments

@avinash Here &a means logical address bcoz of security matter ...
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edited by
it should be option C ?
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Wrong because child and parent are stored in separate memory locations. V can't be equal to Y.
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33 votes
33 votes

answer is c because when child is created, it gets a saparate address space i.e. space in RAM and it clones entire address space of parent into it's own address space. so obviously the variable 'a' is also copied and physical address i.e. location in RAM would be different for both variables. but given program doesn't print the physical address, it prints the logical address i.e. distance between the location of variable in RAM and location of start of address space for the process i.e. relative location of variable with respect to start of it's address space which will be same for both parent and child since data is cloned.

 

in unix tho, address space of parent and child is also same until one of them tries to modify the contents of it's address space. then they are assigned different address spaces and cloning is done.

now before you go on to solve next question, you deserve to see this bunny with a backpack--

edited by

3 Comments

As you said the given program doesn't print the physical address, can you please explain why? I am not getting it
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good explanation🐰🐇
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thanks for the bunny!
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11 votes
11 votes
fork() returns 0 in child process and process ID of child process in parent process.
In Child (x), a = a + 5
In Parent (u), a = a – 5;

Child process will execute the if part and parent process will execute the else part. Assume that the initial value of a = 6. Then the value of a printed by the child process will be 11, and the value of a printed by the parent process in 1. Therefore u+10=x.

the virtual address is same but virtual addresses exist in different processes’ virtual address space and when we print &a, it’s actually printing the virtual address. Hence the answer is v = y.
Answer:

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