GATE CSE 2023 | Question: 25

Question

The integer value printed by the $\textsf{ANSI-C}$ program given below is _______________

#include<stdio.h>

int funcp(){
    static int x = 1;
    x++;
    return x;
}

int main(){
    int x,y;
    x = funcp();
    y = funcp()+x;
    printf("%d\n", (x+y));
    return 0;
}

Comments

In main function the defined x is the local so it won't change

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This helps us understand static storage class.

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Answer 1

So to answer these type of question I like to take some help from Stack Area and Static Area of Main Memory.

We have $3$ areas in Main Memory:

• Stack Area: For storing Local variable.
• Static Area: For storing Global and Static variables.
• Heap Area: For Dynamic Memory Allocation.

In these type of question we are using Stack and Static Area.

After analyzing the code very carefully put Static and Global variables in the Static Area and and auto variables in Stack area.

For the following code:

 

Now in the main function we are calling $funcp()$ function, so activation record of $funcp()$ will be created i.e. a new block named $funcp()$ will be added on the top of $main()$ block in stack. And control will be transfered to $funcp()$.

Now in $funcp()$ function Stack Area’s $x$ will be incremented by one and $2$ will be returned by $funcp()$ function to the $main()$. After that, the activation record of $funcp()$ will be deleted from the stack and control will be transfered to $main()$. Then $2$ will be assigned to $x$(local to main).

Now again $funcp()$ will be called for $y=funcp() + x$ in the main and activation record of $funcp()$ will be created in the stack and the control will be transfered to $funcp()$. After that stack area’s $x$ will be incremented by one and its value will be $3$ and this $3$ will be returned to $main()$ and $y=funcp() + x$ becomes $y = 3 + 2 (funcp() = 3, x = 2)$. Then activation record will be deleted for $funcp()$ and control is transfered to $main()$.

Now we are in main and here $x\text{(local to main)}=2 \text{ and } y=5$, now remember this when global or static & local variables has same name priority is always given to local variable. So in $\text{printf("%d\n", (x+y));}$
 we are accessing local $x$ and local $y$. Hence output will be $x+y = 2+5 = 7$

Answer will be $7$.

Comments

Instead of saying "stack areas x" say "stack areas x stored in static memory"

Answer 2

when $x=funcp()$ is called first time it will go to the function funcp() where local variable $x=1$ is define, increment it and return  value $2$ to local variable $x$ inside main().

when $y=funcp()+x$ is called it will again go to the function funcp(). Since $x$ is a static variable here so new value of $x$ is $2$ itself.increment it and return $3$.

So $y=funcp()+x$ became $y=3+2=5$.

$\therefore x=2,y=5$

$print(x+y)=5+2=7$

the correct answer is $7.$

Answer 3

In main(),

$x = funcp()  →$ makes a call to $funcp() → $ static $int$ $x = 1 , x++ $ changes the value of static $x$ to $2 $ and returns $2 => $ In main, $x=2$ ($x$ is assigned with the value $2$)

$y = funcp() + x  =>     funcp() + 2    $      |---------  makes a call to $funcp() →$  static $int $ $x = 1,$ will not get executed again and again $=> x = 2$ (from previous operations on static $int$ $x$), so $x++$ changes the  value of static $int$ $x$ from $2$ to $3$ and returns $3$ to the caller

$=>$ In main, $ y = 3 + 2 = 5 $

$=> x=2$ and $y=5 $  

$\text{=> printf(“%d\n”, (x+y))}$ prints the decimal value of ${x+y = 2+5 =7}$

Therefore, the output is $7$

Answer 4

Answer:7

Let's break down the code step by step:

1. The program includes the standard input-output library stdio.h.

2. It defines a function funcp() which returns an integer. Inside funcp(), there's a static variable x initialized to 1. Each time funcp() is called, x is incremented by 1 and then returned.

3. In the main() function:

   • It declares two integer variables x and y.
   • Calls funcp() and assigns its return value to x.
   • Calls funcp() again, adds its return value with x, and assigns the result to y.
   • Prints the sum of x and y.

Let's trace the execution:

• First call to funcp(): x becomes 2.
• Second call to funcp(): x becomes 3 (since it's static, its value persists across function calls), and y becomes 3 + 2 = 5.

So, the final output will be the sum of x and y, which is 2 + 5 = 7.

Answer 5

Final Answer is 7

Answer 6

 

1. The Function funcp()

Inside this function, x is static.

• First Call: The computer initializes x = 1 (this happens only once). It then increments x to 2 and returns it.
• Second Call: The computer skips the static int x = 1 line. It sees x is already 2, increments it to 3, and returns it.

The main() Execution

In main, we have local variables x and y. These are separate from the function's x.

• x = funcp();
  • The function returns 2.
  • Main's local x is now 2.
• y = funcp() + x;
  • funcp() is called again. Because it is static, it remembers it was at 2, so it returns 3.
  • We add the local x from main (which is 2).
  • y = 3 + 2(Local variable in main function)  = 5.
• printf("%d\n", (x + y));
  • We add local x (2) and local y (5).
  • 2 + 5 = 7.

The Final Answer is 7.

If x wasn't static, the function would "reset" to 1 every time. Because it is static, the "initialize once" rule allows the value to keep growing.