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: 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.

Q: How do you free allocated memory?

Correct answer: free(var-name);free() returns memory obtained from malloc, calloc or realloc to the heap. delete is the C++ operator, and the other two names do not exist. Freeing the same block twice, or using a pointer after freeing it, is undefined behaviour.

Q: What is the output of this program? #include <stdio.h> void main() { struct demo{ char * a; int n; }; struct demo p = {"hello", 2015}; struct demo q = p; printf("%d", printf("%s",q.a)); }

Correct answer: hello5The inner printf runs first, writing hello and returning 5 - the number of characters it wrote. The outer printf then prints that return value. So the output is hello5, in that order. (The assignment q = p copies the structure, including the pointer, so both point at the same string.)

Q: What will be the output of the program if it is executed like below? cmd> sample /* sample.c */ #include <stdio.h> int main(int argc, char **argv) { printf("%s\n", argv[argc-1]); return 0; }

Correct answer: sampleargv[0] is the program name as invoked, and with no further arguments argc is 1 - so argv[argc-1] is argv[0], which prints sample. The array always has argc entries plus a null pointer at argv[argc].

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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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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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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How do you free allocated memory?

A.remove(var-name);
B.free(var-name);
C.delete(var-name);
D.dalloc(var-name);

Answer: B

Correct answer: free(var-name);

free() returns memory obtained from malloc, calloc or realloc to the heap. delete is the C++ operator, and the other two names do not exist. Freeing the same block twice, or using a pointer after freeing it, is undefined behaviour.

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The above C declaration define 's' to be

struct node 
{ 
   int i; 
   float j; 
}; 
struct node *s[10];

A.An array, each element of which is a pointer to a structure of type node
B.A structure of 2 fields, each field being a pointer to an array of 10 elements
C.A structure of 3 fields: an integer, a float, and an array of 10 elements
D.An array, each element of which is a structure of type node.

Answer: A

Correct answer: An array, each element of which is a pointer to a structure of type node

Subscript binds tighter than the dereference, so struct node *s[10] groups as struct node *(s[10]) - an array of 10 pointers, each to a struct node. A pointer to an array would need parentheses in the other place: struct node (*s)[10].

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What is the output of this program?

#include <stdio.h>         
void main()
{
struct demo{
    char * a;
    int n;
};

struct demo p = {"hello", 2015};
struct demo q = p;
printf("%d", printf("%s",q.a));
}

A.hello
B.5hello
C.hello5
D.6hello

Answer: C

Correct answer: hello5

The inner printf runs first, writing hello and returning 5 - the number of characters it wrote. The outer printf then prints that return value. So the output is hello5, in that order. (The assignment q = p copies the structure, including the pointer, so both point at the same string.)

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10.

What will be the output of the program if it is executed like below?
cmd> sample

/* sample.c */
#include <stdio.h>

int main(int argc, char **argv)
{
    printf("%s\n", argv[argc-1]);
    return 0;
}

A.sample
B.samp
C.No output

Answer: A

Correct answer: sample

argv[0] is the program name as invoked, and with no further arguments argc is 1 - so argv[argc-1] is argv[0], which prints sample. The array always has argc entries plus a null pointer at argv[argc].

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