Consider the following code fragment: if (fork() == 0) { a = a + 5;…

2005

Consider the following code fragment: 

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

Answer: C. u + 10 = x and v = y — Concept: After fork(), the child process is a near-exact duplicate of the parent — both share the same virtual address-space layout, so a given variable has…

  1. A.

    u = x + 10 and v = y

  2. B.

    u = x + 10 and v != y

  3. C.

    u + 10 = x and v = y

  4. D.

    u + 10 = x and v != y

Attempted by 124 students.

Show answer & explanation

Correct answer: C

Concept: After fork(), the child process is a near-exact duplicate of the parent — both share the same virtual address-space layout, so a given variable has the SAME virtual address in parent and child. However, from the instant fork() returns, each process holds its own independent copy of that variable’s value: a write in one process never affects the other’s copy. fork() returns 0 in the child and the child’s process ID in the parent.

Application: Trace this fragment using that rule.

  1. Since fork() returns 0 only in the child, the child takes the if-branch (a = a + 5) and the parent takes the else-branch (a = a - 5).

  2. Let A0 be the value of a immediately before fork() returns (both processes start from this same value).

  3. Child: a becomes A0 + 5, so the child prints x = A0 + 5.

  4. Parent: a becomes A0 - 5, so the parent prints u = A0 - 5.

  5. Subtracting, x - u = (A0 + 5) - (A0 - 5) = 10, i.e. u + 10 = x — this holds for every value of A0.

  6. Both branches also print &a. Per the Concept, the two processes retain the same virtual address for a, so the value printed for &a is identical in both: v = y.

Cross-check: Take a concrete value, say A0 = 20. The parent computes u = 20 - 5 = 15 and the child computes x = 20 + 5 = 25. Check: u + 10 = 15 + 10 = 25 = x — the relation holds, and it does so regardless of which value A0 actually is, so no additional information about a’s initial value is needed.

Conclusion: u + 10 = x and v = y.

Note: printing a pointer with %d (rather than %p) is technically undefined behavior in C; however, this is the exact code as set in the original exam, and every archived derivation of this question treats the value printed for &a as the address value, so the relation v = y is the intended and universally accepted reading.

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