Structures, Unions, Enums Questions and Answers

Q: Consider the following C program.#include\(\lt\)stdio.h\(\gt\)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:

The program prints 0, c.The structure holds three chars in order, x = '1', y = '0' and z = 'a'+2. Since 'a'+2 is evaluated as a character code, z holds 'c'. Casting q to char* and adding 1 and 2 points at y and z, so the two values printed are '0' and 'c'.The format %c prints just the character, without quotes, so the options with quotes are wrong. 'a+2' would need the expression to be stored as text, but it is computed when the structure is initialised.

Q: How much memory will be allocated for the instances of structure S (with respect to C language) ?typedef struct { char a; int b; } S;

An instance of S is allocated the size required to store a character and an integer.A structure stores all of its members at once, each with its own memory, so S holds both a and b. On a typical system with 4-byte integers the compiler also inserts padding after the char to align the int, so sizeof(S) is usually 8 rather than 5.Storing only a character or only an integer would describe a single member; the size of the largest member alone is what a union uses. The size of a structure is fixed at compile time and does not depend on available memory.

Q: What will be the size of the following structure? struct demo{ int a; char b; float c; }

Correct answer: 12Members are laid out with padding so that each is properly aligned:int a4 byteschar b1 byte + 3 paddingfloat c4 bytestotal12 bytesThe three padding bytes exist so that the float begins on a 4-byte boundary. Without alignment the total would be 9.

Q: Which of the following operators can be applied on structure variables?

Correct answer: Assignment ( = )A structure may be assigned as a whole, which copies it member by member. It may not be compared with ==: the compiler could not simply compare the bytes, because padding between members holds unspecified values, and comparing member by member is left to the programmer.

Q: How many bytes does the structure below occupy? #include <stdio.h> struct test { int k; char c; };

Correct answer: Multiple of integer sizeThe members are an int and a char, but the compiler adds padding so that the whole structure aligns to its strictest member. The size therefore comes out a multiple of the int size - 8 bytes on a typical machine, not the 5 the members occupy. Padding is what makes sizeof a structure unpredictable from its members alone.

Q: What will be the output of the following C code? #include <stdio.h> #define MAX 4 enum sanfoundry { a,b=3,c }; main() { if(MAX!=c) printf("hello"); else printf("welcome"); }

Correct answer: welcomeThe enumerators are a = 0, b = 3 and therefore c = 4 - values continue from the last one given. MAX is also 4, so the condition MAX != c is false and the else branch prints welcome.

Consider the following C program.

#include\(\lt\)stdio.h\(\gt\)
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'

Exam Year(s)

GATE CS - 2018,

Answer: A

The program prints 0, c.

The structure holds three chars in order, x = '1', y = '0' and z = 'a'+2. Since 'a'+2 is evaluated as a character code, z holds 'c'. Casting q to char* and adding 1 and 2 points at y and z, so the two values printed are '0' and 'c'.

The format %c prints just the character, without quotes, so the options with quotes are wrong. 'a+2' would need the expression to be stored as text, but it is computed when the structure is initialised.

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How much memory will be allocated for the instances of structure S (with respect to C language) ?

typedef struct {
    char a;
    int b;
} S;

A.size required to store a character
B.size required to store an integer
C.size required to store a character and an integer
D.allocated memory size will depend upon the available memory

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

An instance of S is allocated the size required to store a character and an integer.

A structure stores all of its members at once, each with its own memory, so S holds both a and b. On a typical system with 4-byte integers the compiler also inserts padding after the char to align the int, so sizeof(S) is usually 8 rather than 5.

Storing only a character or only an integer would describe a single member; the size of the largest member alone is what a union uses. The size of a structure is fixed at compile time and does not depend on available memory.

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Which operator is used to access a structure member through a pointer ?

A.->
B..
C.::
D.*

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

A structure member is reached through a pointer with the -> (arrow) operator.

If p points to a structure, p->member is shorthand for (*p).member: it dereferences the pointer and selects the member in one step.

The dot operator is used with a structure variable itself, not a pointer to one. :: is the C++ scope resolution operator, and * on its own only dereferences the pointer; combined with the dot it needs brackets, (*p).member, because . binds more tightly than *.

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Enumeration is a process of

A.Declaring a set of numbers
B.Sorting a list of strings
C.Assigning a legal values possible for a variable
D.Sequencing a list of operators

Exam Year(s)

UGC NET - 2012,

Answer: C

Enumeration is the process of assigning the legal values possible for a variable: an enumerated type lists the named constants that a variable of that type may take, for example the days of a week.

It is not about declaring a set of numbers in general, sorting a list of strings, or sequencing a list of operators. The values in an enumeration are named integer constants, usually numbered from 0 by default.

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What is the size of the following Union ? Assume that the size of int = 2, size of float = 4, size of char = 1

union tag {
int a;
float b;
char c;
};

Exam Year(s)

UGC NET - 2013,

Answer: B

All members of a union share the same memory, so its size is that of its largest member. Here the members are an int of 2 bytes, a float of 4 bytes and a char of 1 byte, so the largest is 4 bytes.

A size of 2 or 1 is smaller than the float, and 7 is the sum of the sizes, which is the size of a structure and not of a union, if padding is ignored. Option B is correct. The compiler allocates the space of the largest member to the union, and the members take turns to use it, so only one of them is valid at a time.

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What will be the size of the following structure?

struct demo{
  int a;
  char b;
  float c;
}

A.12
B.8
C.10
D.9

Answer: A

Correct answer: 12

Members are laid out with padding so that each is properly aligned:

int a	4 bytes
char b	1 byte + 3 padding
float c	4 bytes
total	12 bytes

The three padding bytes exist so that the float begins on a 4-byte boundary. Without alignment the total would be 9.

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Which of the following operators can be applied on structure variables?

A.Equality comparison ( == )
B.Assignment ( = )
C.Both of the above
D.None of the above

Answer: B

Correct answer: Assignment ( = )

A structure may be assigned as a whole, which copies it member by member. It may not be compared with ==: the compiler could not simply compare the bytes, because padding between members holds unspecified values, and comparing member by member is left to the programmer.

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How many bytes does the structure below occupy?

#include <stdio.h>
struct test
{
  int k;
  char c;
};

A.Multiple of integer size
B.Integer size+character size
C.Depends on the platform
D.Multiple of word size

Answer: A

Correct answer: Multiple of integer size

The members are an int and a char, but the compiler adds padding so that the whole structure aligns to its strictest member. The size therefore comes out a multiple of the int size - 8 bytes on a typical machine, not the 5 the members occupy. Padding is what makes sizeof a structure unpredictable from its members alone.

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What will be the output of the following C code?

#include <stdio.h>
#define MAX 4
enum sanfoundry
{
    a,b=3,c
};
main()
{
    if(MAX!=c)
        printf("hello");
    else
        printf("welcome");
}

A.error
B.hello
C.welcome
D.2

Answer: C

Correct answer: welcome

The enumerators are a = 0, b = 3 and therefore c = 4 - values continue from the last one given. MAX is also 4, so the condition MAX != c is false and the else branch prints welcome.

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10.

What will be the output of the following C code if input given is 2?

#include <stdio.h>
enum day
{
    a,b,c=5,d,e
};
main()
{
    printf("Enter the value for a");
    scanf("%d",a);
    printf("%d",a);
}

A.2
B.3
C.Error

Answer: C

The code shown above results in an error. This is because memory is not allocated for constants in enum. Thereore ampersand(&) cannot be used with them. Also, a value has already been assigned to ‘a’, which cannot be changed.

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