GATE CS 2003 Q89 — Programming in C

To find the output of the program, we need to carefully trace the scope of the variable x (global vs. local) and the parameter passing.

1. Variable Scopes

• Global Scope: int x; is declared globally.
• Function Q(int z): Has no local variable named x, so any reference to x inside Q refers to the global x.
• Function P(int *y): Declares a local variable int x. Any reference to x inside P refers to this local x. The pointer y points to the variable passed from main (which is the global x).

2. Step-by-Step Execution Trace

Step 1: Execution starts in main()

• x = 5; → The global variable x is initialized to $5$.
• P(&x); → Function P is called with the address of the global variable x. Thus, inside P, the pointer y points to the global x (i.e., *y is the global x).

Step 2: Inside P(int *y)

• int x = *y + 2; → A local variable x is created. Since *y is the global x (which is $5$), local x becomes $5 + 2 = 7$.
• Q(x); → Calls function Q with the value of local x ($7$) as the argument.

Step 3: Inside Q(int z) with z = 7

• z += x; → Since there is no local x in Q, it uses the global x (which is still $5$). Thus, z becomes $7 + 5 = 12$.
• print(z); → Prints 12.

Step 4: Returning to P

• *y = x - 1; → Updates the value pointed to by y (which is the global x). The local x is $7$, so global x becomes $7 - 1 = 6$.
• print(x); → Prints the local x, which is 7.

Step 5: Returning to main()

• print(x); → Prints the global x, which is now 6.

Conclusion

The sequence of printed values is 12 7 6, which corresponds to Option A.