GATE CSE 2018 | Question: 2

Question

Consider the following C program:

#include<stdio.h>
struct Ournode{
    char x, y, z;
};
int main() {
    struct Ournode p={'1', '0', 'a'+2};
    struct Ournode *q=&p;
    printf("%c, %c", *((char*)q+1), *((char*)q+2));
    return 0;
}

The output of this program is:

• A. 0, c
• B. 0, a+2
• C. '0', 'a+2'
• D. '0', 'c'

Comments

Suppose this code was given  i tested using runing the code in compiler but i dont trust the compiler 

sir can u explain me little bit in depth ::

#include <stdio.h>
struct ournode {
char x,y,z;
};
int main()
{
struct ournode p={'1','0','a'+2};
struct ournode *q=&p;
printf("%c",*(short *)q);
printf("%c",*((short *)q+1));
return 0;
}

 

 

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https://youtu.be/yhzcAfE7LmQ?si=_D5gOrLfA7U0xBTz

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Option d is incorrect because the %c format specifier prints the raw character values themselves, without any surrounding single quotes.

Answer 1

char $x =  $'$a$' $+ 2$
i.e. $x ==$'$c$'

So, $p=${'$1$','$0$','$c$'}

$*((char*)q+1) == p[1]$
$*((char*)q+2) == p[2]$

printf("%c, %c", *((char*)q+1), *((char*)q+2));

Output:  $0, c$

Correct Answer: $A$

Comments

+ has greater precedence than = so 

x = 'a' + 2 --> x = ('a' + 2) --->x = 'c' 

but you interpreted  identifier(variable name) x as character 'x' and (x = 'a') + 2---> x + 2 -->'z'  which is wrong.

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@Digvijay Pandey

What do you mean by $p[1]$ and $p[2]$? I do not quite get it. Please can you elaborate your intention?

Note: That you can initialize a structure in the same way as you can initialize an array does not mean that you can access the elements of structure in the same way.

Doing something as $p[1]$ in C is illegal :

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Yes, this gave error because we already have predefined operators like dot (.) and arrow operator (→) specifically for accessing the structure members.

Answer 2

Important to understand:-

1)

int *p = &a;

In this declaration, p will store the address of a & it only says that the expected value stored at 'a' is integer data type or in simple words, p is pointing to an address which contains integer data. But in practice, a could be anything (int, char, even it could be a structure).    Try it yourself-

int a=101;

char *p=&a;

printf("%d",*p);           output:- 101

2)

WHAT IS CHAR DATA TYPE:- A single letter of any kind is character whether it is a, b, 1, 5, %, $...

only ASCII value of the corresponding letter is stored at the given address. So single integer values are also allowed.

How to store a character into char data type. char a='f'; (by using single quote) & string using double quote("C-programming")

Now come back to question

LINE 1     struct Ournode p = {'1', '0', 'a'+2} ;
LINE 2     struct Ournode *q = &p ;
LINE 3     printf ("%c, %c", *((char*)q+1), *((char*)q+2)) ;

Here pointer q is defined as it is expecting an address of a structure of type ournode. So if we replace LINE 3 with this statement 

printf("%c",*q);                 It will print 1st char data stored in the structure but if we replace LINE 3 with this

printf("%c",*(q+1));           it will jump to next record(skipping structure p completely) & won't print anything, because expected data of pointer q is structure so size of "struct Ournode" will be added to the address stored into pointer q.

Now replace LINE 3 with 

printf ("%c", *((char*)q+1));

if we (Typecast) or change the expected type of the pointer p to char data type then it will not jump to the next record but will add the size of the char data to the address stored into pointer q. so "0" will be printed.

Option A is correct.

If we change LINE 2 & 3 with this...

char *q = &p;

printf ("%c, %c", *(q+1), *(q+2)) ;

then we don't even have to typecast the pointers, It will also work and produces the same result.

 

Comments

char *q = &p;

printf ("%c, %c", *(q+1), *(q+2)) ;

in the first line, implicit typecasting is done right?

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Worth reading !!

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When there is both character and integer remember if you assign integer and character then whole struct take size of integer means every slot wheather character or integer it will take bigger size that means of integer so that is why sometime struct waste memory but this will not happen in union union take mem size as requred like for character only 1B and for integer 4B.

#include<stdio.h>
struct p
{
    int i;
    char j;
};

void main()
{
  struct p c = {1,'A'+2} ;
  struct p *d = &c;
  printf(" %c   %d \n",*((char*)d+sizeof(int)),*((char*)d));  
  return ;
}

Output will be:-   C    1

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That was pin-point explanation!!

Answer 3

Single quotes are used for assigning values to char.

Double quotes are used for assigning values to string.

So, whatever single literal is enclosed in single quotes, will be the value of the char.

 

Hence, 0 and c. NOT '0' and 'c'.

Option A

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To print chars enclosed in single quotes, we'll have to use ' ' explicitly in the print statement.

Answer 4

Answer is: 0,c

char x = 'a' + 2 (ASCII value of a + 2  , so c will be stored)

So, p={'1','0',c};

*((char*)q+1) == p[1]  // q will be character pointer and point to p, which will be incremented by 1. Hence p[1]
*((char*)q+2) == p[2]; // q will be character pointer and point to p, which will be incremented by 2. Hence p[2]
 

Check compiled code here: https://ide.geeksforgeeks.org/jBd3XcMfAX

Answer 5

Answer is: (A) 0,c

char x = 'a' + 2 (ASCII value of a + 2  , so c will be stored)

So, p={'1','0',c};

*((char*)q+1) == p[1]  // q will be character pointer and point to p, which will be incremented by 1. Hence p[1]
*((char*)q+2) == p[2]; // q will be character pointer and point to p, which will be incremented by 2. Hence p[2]

Comments

are the question numbers correct?? I mean 1 to 25-1 marks and 26 to 55-2 marks. So was this question asked only for 1 mark?

Answer 6

'a' + 2 will give 'c'

Ournode p={'1', '0', 'c'}

$*((char*)q+1) \Rightarrow 0$

$*((char*)q+2) \Rightarrow c$

Comments

when it will print with quotes?? what should be the code for that to print in quotes?

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Here q=&p does it mean p also has address and q contains it