C Programming Test

C Programming Mock Test

Free C Programming Mock Test: Placement

Practice the free Online C Programming Test to get success in future placement. This online mock test will help you in the time of final placements.

General Instructions:

1. Each Programming test contains 15 questions.
2. Total allotted time is 20 minutes for each test.
3. No negative marking.
4. Try to finish it on time.
5. The Test will be submitted automatically if the time expired.
6. Don’t refresh the page.

Time is Over

C Programming Test 1

1 / 15

The modulus operator cannot be used with a long double.

A) True

B) False

Explanation:

fmod(x,y) – Calculates x modulo y, the remainder of x/y.
This function is the same as the modulus operator. But fmod() performs floating point or long double divisions

2 / 15

In a function two return statements should never occur.

A) Yes

B) No

Explanation:

No, In a function two return statements can occur but not successively.

Example:

#include <stdio.h>

int mul(int, int); /* Function prototype */

int main()

{

int a = 0, b = 3, c;

c = mul(a, b);

printf(“c = %d\n”, c);

return 0;

}

/* Two return statements in the mul() function */

int mul(int a, int b)

{

if(a == 0 || b == 0)

{

return 0;

}

else

{

return (a * b);

}

Output:
c = 0

3 / 15

How many bytes are occupied by near, far and huge pointers (DOS)?

A) near=2 far=4 huge=4

B) near=4 far=8 huge=8

C) near=2 far=4 huge=8

D) near=4 far=4 huge=8

Explanation:

near=2, far=4 and huge=4 pointers exist only under DOS. Under windows and Linux every pointers is 4 bytes long.

4 / 15

Point out the error in the program?

#include

int main()

{

struct a

{

float category:5;

char scheme:4;

};

printf(“size=%d”, sizeof(struct a));

return 0;

}

A) Error: invalid structure member in

Printf

B) Error in this

float category:5

; statement

C) No Error

D) None of Above

Bit field type must be signed int or unsigned int.

The char type: char scheme:4; is also a valid statement.

5 / 15

Can we specify a variable filed width in a scanf() format string?

A) Yes

B) No

Explanation:

In scanf() a * in a format string after a % sign is used for the suppression of assignment. That is, the current input field is scanned but not stored.

6 / 15

If malloc() successfully allocates memory it returns the number of bytes it has allocated.

A) True

B) False

Explanation:

Syntax: void *malloc(size_t size);

The malloc() function shall allocate unused space for an object whose size in bytes is specified by size and whose value is unspecified.

The order and contiguity of storage allocated by successive calls to malloc() is unspecified. The pointer returned if the allocation succeeds shall be suitably aligned so that it may be assigned to a pointer to any type of object and then used to access such an object in the space allocated (until the space is explicitly freed or reallocated). Each such allocation shall yield a pointer to an object disjoint from any other object. The pointer returned points to the start (lowest byte address) of the allocated space. If the space cannot be allocated, a null pointer shall be returned. If the size of the space requested is 0, the behavior is implementation-defined: the value returned shall be either a null pointer or a unique pointer.

7 / 15

Are the following declarations same?

char far *far *scr;

char far far** scr;

A) Yes

B) No

8 / 15

What will be the output of the program?

#include

int main()

{

printf(“%f\n”, sqrt(36.0));

return 0;

}

A) 6.0

B) 6

C) 6.000000

D) Error: Prototype sqrt() not found.

Explanation:

printf(“%f\n”, sqrt(36.0)); It prints the square root of 36 in the float format(i.e 6.000000).

Declaration Syntax: double sqrt(double x) calculates and return the positive square root of the given number.

9 / 15

In a macro call the control is passed to the macro.

A) True

B) False

Explanation:

False, Always the macro is substituted by the given text/expression.

10 / 15

Which of the following function sets first n characters of a string to a given character?

A) strinit()

B) strnset()

C) strset()

D) strcset()

Explanation:

Declaration:

char *strnset(char *s, int ch, size_t n); Sets the first n characters of s to ch

#include <stdio.h>

#include <string.h>

int main(void)

{

char *string = “abcdefghijklmnopqrstuvwxyz”;

char letter = ‘x’;

printf(“string before strnset: %s\n”, string);

strnset(string, letter, 13);

printf(“string after strnset: %s\n”, string);

return 0;

}

Output:

string before strnset: abcdefghijklmnopqrstuvwxyz

string after strnset: xxxxxxxxxxxxxnopqrstuvwxyz

11 / 15

If the size of pointer is 4 bytes then What will be the output of the program ?

#include

int main()

{

char *str[] = {“Frogs”, “Do”, “Not”, “Die”, “They”, “Croak!”};

printf(“%d, %d”, sizeof(str), strlen(str[0]));

return 0;

}

A) 22, 4

B) 25, 5

C) 24, 5

D) 20, 2

Explanation:

Step 1: char *str[] = {“Frogs”, “Do”, “Not”, “Die”, “They”, “Croak!”}; The variable str is declared as an pointer to the array of 6 strings.

Step 2: printf(“%d, %d”, sizeof(str), strlen(str[0]));

sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints ’24’

strlen(str[0])); becomes strlen(Frogs)). Hence it prints ‘5’;

Hence the output of the program is 24, 5

Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).

12 / 15

What will be the output of the program (myprog.c) given below if it is executed from the command line?

cmd> myprog friday tuesday sunday

/* myprog.c */

#include

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

{

printf(“%c”, *++argv[1]);

return 0;

A) r

B) f

C) m

D) y

13 / 15

Point out the error in the program (in Turbo-C).

#include

#define MAX 128

int main()

{

const int max=128;

char array[max];

char string[MAX];

array[0] = string[0] = ‘A’;

printf(“%c %c\n”, array[0], string[0]);

return 0;

}

A) Error: unknown max in declaration/Constant expression required

B) Error: invalid array string

C) None of above

D) No error. It prints A A

Explanation:

Step 1: A macro named MAX is defined with value 128

Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.

Step 3: char array[max]; This statement reports an error “constant expression required”. Because, we cannot use variable to define the size of array.

To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];

Note: The above program will print A A as output in Unix platform.

14 / 15

What will be the output of the program?

#include

int main()

{

int i;

char c;

for(i=1; i<=5; i++) { scanf("%c", &c); /* given input is 'a' */ printf("%c", c); ungetc(c, stdin); } return 0; }

A) aaaa

B) aaaaa

C) Garbage value

D) Error in ungetc statement

Explanation:

for(i=1; i<=5; i++) Here the for loop runs 5 times.

Loop 1:
scanf(“%c”, &c); Here we give ‘a’ as input.
printf(“%c”, c); prints the character ‘a’ which is given in the previous “scanf()” statement.
ungetc(c, stdin); “ungetc()” function pushes character ‘a’ back into input stream.

Loop 2:
Here the scanf(“%c”, &c); get the input from “stdin” because of “ungetc” function.
printf(“%c”, c); Now variable c = ‘a’. So it prints the character ‘a’.
ungetc(c, stdin); “ungetc()” function pushes character ‘a’ back into input stream.

This above process will be repeated in Loop 3, Loop 4, Loop 5.

15 / 15

Which of the following statements are correct about an if-else statements in a C-program?

1	Every if-else statement can be replaced by an equivalent statements using ?: operators
2	Nested if-else statements are allowed.
3	Multiple statements in an if block are allowed.
4	Multiple statements in an else block are allowed.

A) 1 and 2

B) 2 and 3

C) 1, 2 and 4

D) 2, 3, 4

Your score is

The average score is 36%

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C Programming Test 2

1 / 15

A short integer is at least 16 bits wide and a long integer is at least 32 bits wide

A) True

B) False

Explanation:

The basic C compiler is 16 bit compiler, below are the size of it’s data types
The size of short int is 2 bytes wide(16 bits).
The size of long int is 4 bytes wide(32 bits).

2 / 15

In which order do the following gets evaluated

1	Relational
2	Arithmetic
3	Logical
4	Assignment

A) 2134

B) 1234

C) 4321

D) 3214

Explanation:

2. Arithmetic operators: *, /, %, +, –
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, ||
4. Assignment operators: =

3 / 15

What will be the output of the program?

#include

#define SQUARE(x) x*x

int main()

{

float s=10, u=30, t=2, a;

a = 2*(s-u*t)/SQUARE(t);

printf(“Result = %f”, a);

return 0;

}

A) Result = -100.000000

B) Result = -25.000000

C) Result = 0.000000

D) Result = 100.000000

Explanation:

The macro function SQUARE(x) x*x calculate the square of the given number ‘x’. (Eg: 10²)

Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.

Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,

=> a = 2 * (10 – 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .

=> a = 2 * (10 – 30 * 2) / 2 * 2;

=> a = 2 * (10 – 60) / 2 * 2;

=> a = 2 * (-50) / 2 * 2 ;

=> a = 2 * (-25) * 2 ;

=> a = (-50) * 2 ;

=> a = -100;

Step 3: printf(“Result=%f”, a); It prints the value of variable ‘a’.

Hence the output of the program is -100

4 / 15

A float occupies 4 bytes. If the hexadecimal equivalent of these 4 bytes are A, B, C and D, then when this float is stored in memory in which of the following order do these bytes gets stored?

A) ABCD

B) DCBA

C) 0xABCD

D) Depends on big endian or little endian architecture

5 / 15

What will be the output of the program ?

#include

int main()

{

char str1[20] = “Hello”, str2[20] = ” World”;

printf(“%s\n”, strcpy(str2, strcat(str1, str2)));

return 0;

}

A) Hello

B) World

C) Hello World

D) WorldHello

Explanation:

Step 1: char str1[20] = “Hello”, str2[20] = ” World”; The variable str1 and str2 is declared as an array of characters and initialized with value “Hello” and ” World” respectively.

Step 2: printf(“%s\n”, strcpy(str2, strcat(str1, str2)));

=> strcat(str1, str2)) it append the string str2 to str1. The result will be stored in str1. Therefore str1 contains “Hello World”.

=> strcpy(str2, “Hello World”) it copies the “Hello World” to the variable str2.

Hence it prints “Hello World”.

6 / 15

What will be the output of the program ?

#include

int main()

{

union var

{

int a, b;

};

union var v;

v.a=10;

v.b=20;

printf(“%d\n”, v.a);

return 0;

}

A) 10

B) 20

C) 30

D) 0

7 / 15

Can we have an array of bit fields?

A) Yes

B) No

8 / 15

What will be the output of the program?

#include

int main()

{

unsigned int res;

res = (64 >>(2+1-2)) & (~(1<<2)); printf(“%d\n”, res);

return 0;

}

A) 32

B) 64

C) 0

D) 128

9 / 15

Which of the following cannot be checked in a switch-case statement?

A) Character

B) Integer

C) Float

D) enum

Explanation:

The switch/case statement in the c language is defined by the language specification to use an int value, so you can not use a float value.

switch( expression )

{

case constant-expression1: statements 1;

case constant-expression2: statements 2;

case constant-expression3: statements3 ;

…

default : statements 4;

}

The value of the ‘expression’ in a switch-case statement must be an integer, char, short, long. Float and double are not allowed.

10 / 15

Which of the following statements are correct about the program?

#include

int main()

{

int x = 30, y = 40;

if(x == y)

printf(“x is equal to y\n”);

else if(x > y)

printf(“x is greater than y\n”);

else if(x < y) printf(“x is less than y\n”)

return 0;

}

A) Error: Statement missing

B) Error: Expression syntax

C) Error: Lvalue required

D) Error: Rvalue required

Explanation:

This program will result in error “Statement missing ;”

printf(“x is less than y\n”) here ; should be added to the end of this statement.

11 / 15

Macros have a local scope

A) True

B) False

Explanation:

False, The scope of macros is globals and functions. Also the scope of macros is only from the point of definition to the end of the file.

12 / 15

Is this a correct way for NULL pointer assignment?

int i=0;
char *q=(char*)i;

A) Yes

B) No

Explanation:

The correct way is char *q=0 (or) char *q=(char*)0

13 / 15

What will be the output of the program ?

#include

int main()

{

char sentence[80];

int i;

printf(“Enter a line of text\n”);

gets(sentence);

for(i=strlen(sentence)-1; i >=0; i–)

putchar(sentence[i]);

return 0;

}

A) The sentence will get printed in same order as it entered

B) The sentence will get printed in reverse order

C) Half of the sentence will get printed

D) None of above

14 / 15

Point out the error in the program?

struct emp

{

int ecode;

struct emp *e;

};

A) Error: in structure declaration

B) Linker Error

C) No Error

D) None of above

Explanation:

This type of declaration is called as self-referential structure. Here *e is pointer to a struct emp.

15 / 15

In Turbo C/C++ under DOS if we want that any wild card characters in the command-line arguments should be appropriately expanded, are we required to make any special provision?

A) Yes

B) No

Explanation:

Yes you have to compile a program like
tcc myprog wildargs.obj

Your score is

The average score is 38%

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C Programming Test 3

1 / 15

We can modify the pointers “source” as well as “target”.

A) True

B) False

2 / 15

What do the following declaration signify?

void *cmp();

A) cmp is a pointer to an void type

B) cmp is a void type pointer variable.

C) cmp is a function that return a void pointer.

D) cmp function returns nothing.

3 / 15

What will be the output of the program?

#include

int get();

int main()

{

const int x = get();

printf(“%d”, x);

return 0;

}

int get()

{

return 20;

}

A) Garbage value

B) Error

C) 20

D) 0

Explanation:

Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.

Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value “20”.

The function get() returns the value “20”.

Step 3: printf(“%d”, x); It prints the value of the variable x.

Hence the output of the program is “20”.

4 / 15

Which of the following statements correct about the below program?

#include

int main()

{

union a

{

int i;

char ch[2];

};

union a u1 = {512};

union a u2 = {0, 2};

return 0;

}

1	u2 CANNOT be initialized as shown.
2	u1 can be initialized as shown.
3	To initialize char ch[] of u2 ‘.’ operator should be used.
4	The code causes an error ‘Declaration syntax error’

A) 1, 2

B) 2, 3

C) 1, 2, 3

D) 1, 3, 4

5 / 15

Is there any difference int the following declarations?
int fun(int arr[]);
int fun(int arr[2]);

A) Yes

B) No

Explanation:

No, both the statements are same. It is the prototype for the function fun() that accepts one integer array as an parameter and returns an integer value.

6 / 15

Point out the error in the program

#include

int main()

{

int i;

#if A

printf(“Enter any number:”);

scanf(“%d”, &i);

#elif B

printf(“The number is odd”);

return 0;

}

A) Error: unexpected end of file because there is no matching #endif

B) The number is odd

C) Garbage values

D) None of above

Explanation:

The conditional macro #if must have an #endif. In this program there is no #endif statement written.

7 / 15

What will be the output of the program?

#include

int main()

{

float n=1.54;

printf(“%f, %f\n”, ceil(n), floor(n));

return 0;

}

A) 2.000000, 1.000000

B) 1.500000, 1.500000

C) 1.550000, 2.000000

D) 1.000000, 2.000000

Explanation:

ceil(x) round up the given value. It finds the smallest integer not < x.
floor(x) round down the given value. It finds the smallest integer not > x.

printf(“%f, %f\n”, ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1.

In the printf(“%f, %f\n”, ceil(n), floor(n)); statement, the format specifier “%f %f” tells output to be float value. Hence it prints 2.000000 and 1.000000.

8 / 15

Point out the correct statements are correct about the program below?

#include

int main()

{

char ch;

while(x=0;x<=255;x++) printf(“ASCII value of %d character %c\n”, x, x);

return 0;

}

A) The code generates an infinite loop

B) The code prints all ASCII values and its characters

C) Error: x undeclared identifier

D) Error: while statement missing

Explanation:

There are 2 errors in this program.
1. “Undefined symbol x” error. Here x is not defined in the program.
2. Here while() statement syntax error.

9 / 15

What will be the output of the program?

#include

int main()

{

const c = -11;

const int d = 34;

printf(“%d, %d\n”, c, d);

return 0;

}

A) Error

B) -11, 34

C) 11, 34

Explanation:

Step 1: const c = -11; The constant variable ‘c’ is declared and initialized to value “-11”.

Step 2: const int d = 34; The constant variable ‘d’ is declared as an integer and initialized to value ’34’.

Step 3: printf(“%d, %d\n”, c, d); The value of the variable ‘c’ and ‘d’ are printed.

10 / 15

What will be the output of the program?

#include

int main()

{

const c = -11;

const int d = 34;

printf(“%d, %d\n”, c, d);

return 0;

}

A) Error

B) -11, 34

C) 11, 34

D) None of these

11 / 15

The program will generate infinite loop. When an EOF is encountered fgetc()returns EOF. Instead of checking the condition for EOF we have checked it for NULL. so the program will generate infinite loop.

The maximum combined length of the command-line arguments including the spaces between adjacent arguments is

A) 128 characters

B) 256 characters

C) 67 characters

D) It may vary from one operating system to another

12 / 15

What will be the output of the program ?

#include

int main()

{

FILE *ptr;

char i;

ptr = fopen(“myfile.c”, “r”);

while((i=fgetc(ptr))!=NULL)

printf(“%c”, i);

return 0;

}

A) Print the contents of file "myfile.c"

B) Print the contents of file "myfile.c" upto NULL character

C) Infinite loop

D) Error in program

13 / 15

What will be the output of the program ?

#include

int main()

{

enum status {pass, fail, absent};

enum status stud1, stud2, stud3;

stud1 = pass;

stud2 = absent;

stud3 = fail;

printf(“%d %d %d\n”, stud1, stud2, stud3);

return 0;

}

A) 0, 1, 2

B) 1, 2, 3

C) 0, 2, 1

D) 1, 3, 2

14 / 15

Will the printf() statement print the same values for any values of a ?

#include

int main()

{

float a;

scanf(“%f”, &a);

printf(“%f\n”, a+a+a);

printf(“%f\n”, 3*a);

return 0;

}

A) Yes

B) No

15 / 15

Which of the following is the correct order of evaluation for the below expression?
z = x + y * z / 4 % 2 – 1

A) * / % + – =

B) = * / % + –

C) / * % – + =

D) * % / – + =

Explanation:

C uses left associativity for evaluating expressions to break a tie between two operators having same precedence

Your score is

The average score is 37%

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C Programming Test 4

1 / 15

We can modify the pointers “source” as well as “target”.

A) True

B) False

2 / 15

What will be the output of the program (myprog.c) given below if it is executed from the command line?
cmd> myprog one two three

/* myprog.c */

#include

int main(int argc, char **argv)

{

printf(“%s\n”, *++argv);

return 0;

}

A) myprog

B) one

C) two

D) three

3 / 15

In a file contains the line “I am a boy\r\n” then on reading this line into the array str using fgets(). What will str contain?

A) "I am a boy\r\n\0"

B) "I am a boy\r\0"

C) "I am a boy\n\0"

D) "I am a boy"

Explanation:

Declaration: char *fgets(char *s, int n, FILE *stream);

fgets reads characters from stream into the string s. It stops when it reads either n – 1 characters or a newline character, whichever comes first.

Therefore, the string str contain “I am a boy\n\0”

4 / 15

A preprocessor directive is a message from programmer to the preprocessor

A) True

B) False

Explanation:

True, the programmer tells the compiler to include the preprocessor when compiling.

5 / 15

In the expression a=b=5 the order of Assignment is NOT decided by Associativity of operators

A) True

B) False

Explanation:

The equal to = operator has Right-to-Left Associativity. So it assigns b=5 then a=b.

6 / 15

What will be the output of the program?

#include

int main()

{

int i=2;

printf(“%d, %d\n”, ++i, ++i);

return 0;

}

A) 3, 4

B) 4, 3

C) 4, 4

D) Output may vary from compiler to compiler

Explanation:

The order of evaluation of arguments passed to a function call is unspecified.

Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.

In TurboC, the output will be 4, 3.

In GCC, the output will be 4, 4.

7 / 15

How many times the while loop will get executed if a short int is 2 byte wide?

#include

int main()

{

int j=1;

while(j <= 255) { printf(“%c %d\n”, j, j);

j++;

}

return 0;

}

A) Infinite times

B) 255 times

C) 256 times

D) 254 times

Explanation:

The while(j <= 255) loop will get executed 255 times. The size short int(2 byte wide) does not affect the while() loop.

8 / 15

va_list is an array that holds information needed by va_arg and va_end

A) True

B) False

9 / 15

The first argument to be supplied at command-line must always be count of total arguments.

A) True

B) False

10 / 15

Bit fields CANNOT be used in union.

A) True

B) False

Explanation:

The following is the example program to explain “using bit fields inside an union”.

#include<stdio.h>

union Point

{

unsigned int x:4;

unsigned int y:4;

int res;

};

int main()

{

union Point pt;

pt.x = 2;

pt.y = 3;

pt.res = pt.y;

printf(“\n The value of res = %d” , pt.res);

return 0;

}

// Output: The value of res = 3

11 / 15

If char=1, int=4, and float=4 bytes size, What will be the output of the program ?

#include

int main()

{

char ch = ‘A’;

printf(“%d, %d, %d”, sizeof(ch), sizeof(‘A’), sizeof(3.14f));

return 0;

}

A) 1, 2, 4

B) 1, 4, 4

C) 2, 2, 4

D) 2, 4, 8

Explanation:

Step 1: char ch = ‘A’; The variable ch is declared as an character type and initialized with value ‘A’.

Step 2:

printf(“%d, %d, %d”, sizeof(ch), sizeof(‘A’), sizeof(3.14));

The sizeof function returns the size of the given expression.

sizeof(ch) becomes sizeof(char). The size of char is 1 byte.

sizeof(‘A’) becomes sizeof(65). The size of int is 4 bytes (as mentioned in the question).

sizeof(3.14f). The size of float is 4 bytes.

Hence the output of the program is 1, 4, 4

12 / 15

Which of the following statements are correct about an array?

1	The array int num[26]; can store 26 elements.
2	The expression num[1] designates the very first element in the array.
3	It is necessary to initialize the array at the time of declaration.
4	The declaration num[SIZE] is allowed if SIZE is a macro.

A) 1

B) 1, 4

C) 2, 3

D) 2, 4

Explanation:

The array int num[26]; can store 26 elements. This statement is true.
The expression num[1] designates the very first element in the array. This statement is false, because it designates the second element of the array.
It is necessary to initialize the array at the time of declaration. This statement is false.
The declaration num[SIZE] is allowed if SIZE is a macro. This statement is true, because the MACRO just replaces the symbol SIZE with given value.

Hence the statements ‘1’ and ‘4’ are correct statements.

13 / 15

What will be the output of the program?

#include

int main()

{

int k, num=30;

k = (num>5 ? (num <=10 ? 100 : 200): 500); printf(“%d\n”, num);

return 0;

}

A) 200

B) 30

C) 100

D) 500

Explanation:

Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to ’30’.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf(“%d\n”, num); It prints the value of variable num ’30’
Step 3: Hence the output of the program is ’30’

14 / 15

The keyword used to transfer control from a function back to the calling function is

A) switch

B) goto

C) go back

D) return

Explanation:

The keyword return is used to transfer control from a function back to the calling function.

Example:

#include<stdio.h>

int add(int, int); /* Function prototype */

int main()

{

int a = 4, b = 3, c;

c = add(a, b);

printf(“c = %d\n”, c);

return 0;

}

int add(int a, int b)

{

/* returns the value and control back to main() function */

return (a+b);

}

Output:
c = 7

15 / 15

Which of the statements is correct about the program?

#include

int main()

{

float a=3.14;

char *j;

j = (char*)&a;

printf(“%d\n”, *j);

return 0;

}

A) It prints ASCII value of the binary number present in the first byte of a float variable a.

B) It prints character equivalent of the binary number present in the first byte of a float variable a.

C) It will print 3

D) It will print a garbage value

Your score is

The average score is 57%

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Time is Over

C Programming Test 5

1 / 15

Which statement will you add in the following program to work it correctly?

#include

int main()

{

printf(“%f\n”, log(36.0));

return 0;

}

A) #include<conio.h>

B) #include<math.h>

C) #include<stdlib.h>

D) #include<dos.h>

Explanation:

math.h is a header file in the standard library of C programming language designed for basic mathematical operations.

Declaration syntax: double log(double);

2 / 15

What will be the output of the program assuming that the array begins at the location 1002 and size of an integer is 4 bytes?

#include

int main()

{

int a[3][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 };

printf(“%u, %u, %u\n”, a[0]+1, *(a[0]+1), *(*(a+0)+1));

return 0;

}

A) 448, 4, 4

B) 520, 2, 2

C) 1006, 2, 2

D) Error

3 / 15

What will be the output of the program in 16-bit platform (Turbo C under DOS) ?

#include

int main()

{

printf(“%d, %d, %d”, sizeof(3.0f), sizeof(‘3’), sizeof(3.0));

return 0;

}

A) 8, 1, 4

B) 4, 2, 8

C) 4, 2, 4

D) 10, 3, 4

Explanation:

Step 1:

printf(“%d, %d, %d”, sizeof(3.0f), sizeof(‘3’), sizeof(3.0));

The sizeof function returns the size of the given expression.

sizeof(3.0f) is a floating point constant. The size of float is 4 bytes

sizeof(‘3’) It converts ‘3’ in to ASCII value.. The size of int is 2 bytes

sizeof(3.0) is a double constant. The size of double is 8 bytes

Hence the output of the program is 4,2,8

Note: The above program may produce different output in other platform due to the platform dependency of C compiler.

In Turbo C, 4 2 8. But in GCC, the output will be 4 4 8.

4 / 15

A structure can contain similar or dissimilar elements

A) True

B) False

5 / 15

What will be the output of the program if value 25 given to scanf()?

#include

int main()

{

int i;

printf(“%d\n”, scanf(“%d”, &i));

return 0;

}

A) 24

B) 2

C) 1

D) 5

Explanation:

The scanf function returns the number of input is given.

printf(“%d\n”, scanf(“%d”, &i)); The scanf function returns the value 1(one).

Therefore, the output of the program is ‘1’.

6 / 15

What will be the output of the program (sample.c) given below if it is executed from the command line (turbo c under DOS)?
cmd> sample Good Morning

/* sample.c */

#include

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

{

printf(“%d %s”, argc, argv[1]);

return 0;

}

A) 3 Good

B) 2 Good

C) Good Morning

D) 3 Morning

7 / 15

Even if integer/float arguments are supplied at command prompt they are treated as strings.

A) True

B) False

8 / 15

Data written into a file using fwrite() can be read back using fscanf()

A) True

B) False

9 / 15

What will be the output of the program?

#include

int main()

{

int i=-3, j=2, k=0, m;

m = ++i && ++j || ++k;

printf(“%d, %d, %d, %d\n”, i, j, k, m);

return 0;

}

A) 1, 2, 0, 1

B) -3, 2, 0, 1

C) -2, 3, 0, 1

D) 2, 3, 1, 1

Explanation:

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns ‘1’(one). Hence m=1.

Step 3: printf(“%d, %d, %d, %d\n”, i, j, k, m); In the previous step the value of i,j are increemented by ‘1’(one).

Hence the output is “-2, 3, 0, 1”.

10 / 15

If the size of integer is 4bytes, What will be the output of the program?

#include

int main()

{

int arr[] = {12, 13, 14, 15, 16};

printf(“%d, %d, %d\n”, sizeof(arr), sizeof(*arr), sizeof(arr[0]));

return 0;

}

A) 10, 2, 4

B) 20, 4, 4

C) 16, 2, 2

D) 20, 2, 2

11 / 15

What will be the output of the program ?

#include

int main()

{

int i;

char a[] = “\0”;

if(printf(“%s”, a))

printf(“The string is empty\n”);

else

printf(“The string is not empty\n”);

return 0;

}

A) The string is empty

B) The string is not empty

C) No output

D) 0

Explanation:

The function printf() returns the number of charecters printed on the console.

Step 1: char a[] = “\0”; The variable a is declared as an array of characters and it initialized with “\0”. It denotes that the string is empty.

Step 2: if(printf(“%s”, a)) The printf() statement does not print anything, so it returns ‘0’(zero). Hence the if condition is failed.

In the else part it prints “The string is not empty”.

12 / 15

If a char is 1 byte wide, an integer is 2 bytes wide and a long integer is 4 bytes wide then will the following structure always occupy 7 bytes?

struct ex

{

char ch;

int i;

long int a;

};

A) Yes

B) No

Explanation:

A compiler may leave holes in structures by padding the first char in the structure with another byte just to ensures that the integer that follows is stored at an location. Also, there might be 2extra bytes after the integer to ensure that the long integer is stored at an address, which is multiple of 4. Such alignment is done by machines to improve the efficiency of accessing values.

13 / 15

What will be the output of the program

#include

void fun(int);

int main(int argc)

{

printf(“%d “, argc);

fun(argc);

return 0;

}

void fun(int i)

{

if(i!=4)

main(++i);

}

A) 1 2 3

B) 1 2 3 4

C) 2 3 4

D) 1

14 / 15

Can I increase the size of dynamically allocated array?

A) Yes

B) No

Explanation:

Use realloc(variable_name, value);

15 / 15

What will be the output of the program (in Turbo C under DOS)?

#include

int main()

{

char huge *near *far *ptr1;

char near *far *huge *ptr2;

char far *huge *near *ptr3;

printf(“%d, %d, %d\n”, sizeof(ptr1), sizeof(ptr2), sizeof(ptr3));

return 0;

}

A) 4, 4, 8

B) 2, 4, 4

C) 4, 4, 2

D) 2, 4, 8

Your score is

The average score is 46%

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