GATE CS 2023 Q25 — Programming in C

To find the output of the program, let's trace the execution step-by-step. Note that the variable x inside funcp() is declared as static, meaning it retains its value across multiple function calls and is initialized only once.

Step-by-Step Execution:

1. First call to funcp():
   • The static variable funcp::x is initialized to $1$ (this initialization happens only once).
   • x++ increments funcp::x to $2$.
   • The function returns $2$.
   • In main(), the local variable x is assigned this returned value: $$\text{main::x} = 2$$
2. Second call to funcp():
   • The static variable funcp::x retains its previous value of $2$.
   • x++ increments funcp::x to $3$.
   • The function returns $3$.
3. Evaluating y = funcp() + x in main():
   • The call funcp() returns $3$.
   • The local variable main::x is $2$.
   • Thus, $$\text{main::y} = 3 + 2 = 5$$
4. Printing the result:
   • The program prints the value of x + y: $$\text{main::x} + \text{main::y} = 2 + 5 = 7$$

Therefore, the integer value printed by the program is 7.