GATE CSE 2003 | Question: 89

Question

Consider the C program shown below:

#include<stdio.h>
#define print(x) printf("%d", x)

int x;
void Q(int z)
{
        z+=x;
        print(z);
}

void P(int *y)
{
        int x = *y + 2;
        Q(x);
        *y = x - 1;
        print(x);
}
main(void) {
        x = 5;
        P(&x);
        print(x);
}

The output of this program is:

• A. $12 \ 7 \ 6$
• B. $22 \ 12 \ 11$
• C. $14 \ 6 \ 6$
• D. $7 \ 6 \ 6$

Comments

In the above question we have declared x as a global variable and then we are initializing x=5 in the activation record of 'main', so shouldn't be it x=5 for the main function and x=0 as a global function ?

---

there is just one global variable x in data segment. same memory location is referred whenever x is encountered in a function. but in case of P function there is a local var named "x", which will override the global variable , and any referrence in this block to x is for its local x.

---

@Shivadu since x has been declared globally (in starting of code) so we can use x=5 locally as well as globally

Answer 1

main: x = 5; //Global x becomes 5

P: int x = *y + 2; //local x in P becomes 5+2 = 7

Q: z+=x; //local z in Q becomes 7 + 5 = 12

Q: print(z); //prints 12

P:  *y  = x - 1; 
//content of address of local variable y 
(same as global variable x) becomes 7 - 1 = 6

P: print(x); //prints local variable x in P = 7

main: print(x); //prints the global variable x = 6

Correct Answer: $A$

Comments

By considering dynamic scoping output will be 14 7 6.

---

@sutanay3 becoz y in P() is being referred by the main fun() which have x and that is the reason it is printing x value as 6 in last 

---

In the line 

1. P: *y = x - 1;
2. //content of address of local variable y
3. (same as global variable x) becomes 7 - 1 = 6

we can say like the local var is inside the p is destroyed because of after the function call so instead of local var 7 we access the 5 global var 

Answer 2

The answer for this is A).. Can get the answer by a simple dry run..

Comments

As in the above program int x is initialized inside the main which is acting as the gloabl value of x?

---

Nopes. Initialization means assigning initial value when memory is created. That happens before main starts execution. In C, if a value is not given, all global/static variables are initialized to 0. So, in the given question x is initialized to 0 and in main the value gets changed to 5.

---

thank you @Arjun sir you made me clear now. :)

Answer 3

 Output is 12 7 6

Answer 4

https://www.youtube.com/watch?v=GEo7asswBhc

Answer 5

void P(int *y)
{
        int x = *y + 2; // x=5+2=7
        Q(x); // Q(7)
        void Q(int z) // z=7 
           { 
               z+=x; // z=z+x=7+5=12 as here x is global and its value is 5 define in main() section print(z); // 12 is printed 
           } 
        *y = x - 1; // 7-1=6
        print(x);  // here value of x is 7 as it is local so 7 is printed   
}                     

 print(x); // here value of y is printed i.e 6 as it represents the final value of global x

so output is 12 7 6

Answer 6

Anwser is in the uploaded image