GATE CS 2012 Q49 — Programming in C

To find the output of the given C code segment, let us trace the execution step-by-step with the specified modifications.

1. Variable Declarations and Scopes

• Global variables:
  int a, b, c = 0;
  Since they are declared globally, they are initialized to 0 by default. Thus, global $a = 0$ and global $b = 0$.
• Inside main():
  Line 1 is replaced by auto int a = 1;. This creates a local variable $a$ inside main(), initialized to $1$.
• Inside prtFun():
  Line 2 is replaced by register int a = 2;. A register variable has local scope and automatic lifetime (similar to auto). It is re-allocated and re-initialized to $2$ every time prtFun() is called.
  int b = 1; is a local automatic variable, re-initialized to $1$ on every call.

2. Step-by-Step Trace

Step 1: First call to prtFun() from main()

• Local variable $a$ (register) is initialized to $2$.
• Local variable $b$ (auto) is initialized to $1$.
• The statement a += ++b; is executed:
  • ++b pre-increments local $b$ from $1$ to $2$.
  • a += 2 updates local $a$ to $2 + 2 = 4$.
• printf(" \n %d %d ", a, b); prints the local variables of prtFun():
  Output: 4 2

Step 2: Return to main() and increment

• a += 1; modifies the local variable $a$ of main().
  Local $a$ becomes $1 + 1 = 2$.

Step 3: Second call to prtFun() from main()

• Since $a$ is a register variable (not static), it does not retain its previous value. It is re-initialized to $2$.
• Local $b$ is re-initialized to $1$.
• The statement a += ++b; is executed:
  • ++b increments local $b$ from $1$ to $2$.
  • a += 2 updates local $a$ to $2 + 2 = 4$.
• printf(" \n %d %d ", a, b); prints:
  Output: 4 2

Step 4: Return to main() and final print

• printf(" \n %d %d ", a, b); is executed inside main():
  • The variable a refers to the local variable $a$ of main(), which has the value $2$.
  • The variable b refers to the global variable $b$ (since there is no local variable $b$ declared in main()), which has the value $0$.
• Output: 2 0

Conclusion

4  2
4  2
2  0