Operating Systems | Process Management | Question 6

Consider the following code fragment:

filter_none

edit
close

play_arrow

link
brightness_4
code

if (fork() == 0)
{ a = a + 5; printf("%d,%d\n", a, &a); }
else { a = a –5; printf("%d, %d\n", a, &a); } 

chevron_right


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?

(A) u = x + 10 and v = y
(B) u = x + 10 and v != y
(C) u + 10 = x and v = y
(D) u + 10 = x and v != y


Answer: (C)

Explanation: 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;
Therefore x = u + 10.
The physical addresses of ‘a’ in parent and child must be different. But our program accesses virtual addresses (assuming we are running on an OS that uses virtual memory). The child process gets an exact copy of parent process and virtual address of ‘a’ doesn’t change in child process. Therefore, we get same addresses in both parent and child. See this run for example.

Quiz of this Question



My Personal Notes arrow_drop_up

Improved By : ArthurDayne05