GATE CSE 2022 | Question: 11

Question

What is printed by the following $\text{ANSI C}$ program?

#include<stdio.h>

int main(int argc, char *argv[])

{

    int x = 1, z[2] = {10, 11};

    int *p = NULL;

    p = &x;

    *p = 10;

    p = &z[1];

    *(&z[0] + 1) += 3;

    printf(“%d, %d, %d\n”, x, z[0], z[1]);

    return 0;

}

• A. $1, 10, 11$
• B. $1, 10, 14$
• C. $10, 14, 11$
• D. $10, 10, 14$

Answer 1

 Option D is correct answer.

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Given that x=1, z[0]=10 and z[1]=11
 
 

int *p=NULL;
p=&x;
*p=10;

By this code, they are accessing x and changing it’s value to 10. Therefore x=10 after the above code executed.

 

*(&z[0]+1) += 3; 

  As per precedence rule, z[0], &z[0], &z[0]+1, *(&z[0]+1) and *(&z[0]+1) += 3; evaluated in the order

&z[0] --- address of z[0]

&z[0] + 1 ---- address of z[1]   $\text{ [ because z[0] and z[1] are contiguously allocated in the memory ]}$

So, we can rewrite the above statement as *( address of z[1] ) += 3

which is nothing but z[1] += 3 ==> z[1] = z[1] + 3

Therefore z[1] = 14

Notice that, z[0] not changed.

Comments

Best

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well explained

Answer 2

1. int x = 1, z[2] = {10, 11};
2. // Initially they are intialized one integer and one 1D array in the above statement//
3. int *p = NULL;
4. // here they initializing one integer pointer and assigning the address as null//
5. p = &x;
6. //here they are assigning address of x to the p //
7. *p = 10;
8. // here they are changing the value as 10 at the address of p is pointing // 
9. Through this statement x will become 10
10. p = &z[1];
11. // now they are assigning address of 2^nd element to the pointer p,by dereferencing previous address//
12. *(&z[0] + 1) += 3;
13. // &z[0] means 1^st element address //
14.   &z[0]+1 means next element of &z[0] i.e 2^nd element in the array
15. *(&z[0]+1) the value present at that address i.e 11(the value of second element in the array)
16. by using assignment operator *(&z[0]+1)=*(&z[0]+1)+3 is equal 11+3=14(it is replaces the second element in the array as 14)
17. printf(“%d, %d, %d\n”, x, z[0], z[1]);
18. Answer: x=10,z[0]=10,z[1]=14.

Answer 3

the correct answer is option “D”

---

now let me explain how?

4^th line is normal, assigning the values to int x and an array “z”

*p is a pointer and initially its value is null or 0. now in the 6^th line, we store or assign the address of “x” into “p”.

now in the pointer, we update the value from null or 0 to 10. since the pointer is pointing to “int x” therefore the value of int x will also change to 10. now we update the pointer to z[1] or {11}. now at last *(&z[0] + 1) += 3 ==> (&z[0]+1 means value at address z[0]+1, which is 11……. 11+3=14) here value of z[1] will update to 14

 

now printf(“%d, %d, %d\n”, x, z[0], z[1]) will print 10,10,14

 

Answer 4

x = 1 , z[2] = { 10 , 11 }

*p = Null ; p = &x; // which means p stores the x adress

*p = 10; // which means x value 10

p = &z[1] // which means p stores the z[1] adress or it can also be written as p = z+1 or &z[1] both are the addresses

Here ,

*( &z[0] + 1 ) += 3;  // we know that the array is in consecutive address, so &z[0] + 1 = &z[1]

*( &z[1] ) += 3; // given that p = &z[1]

*(p) = *(p) + 3;

*(p) = 11 + 3 = 14;

so x = 10, z[0] = 10 not changed , z[1] = 14