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C Language Questions And Answers Sample Test 1


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C Language Test 1


You have scored 3 out of 10
You Scored:2
Total Questions:2
Attended Questions:2
Correct Answered:2
1.
    What will be the output of the program?

#include
#define PRINT(i) printf("%d,",i)

int main()
{
    int x=2, y=3, z=4;
    PRINT(x);
    PRINT(y);
    PRINT(z);
    return 0;
}



Explanation:

The macro PRINT(i) print("%d,", i); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("%d,",x). Hence it prints '2'.

Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'.

Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'.

Hence the output of the program is 2, 3, 4.


2.
What will be the output of the program?

#include<stdio.h>
#define MIN(x, y) (x<y)? x : y;
int main()
{
    int x=3, y=4, z;
    z = MIN(x+y/2, y-1);
    if(z > 0)
        printf("%dn", z);
    return 0;
}



Explanation:

The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.

Step 1: int x=3, y=4, z; The variable x, y, z are declared as an integer type and the variable x, y are initialized to value 3, 4 respectively.

Step 2: z = MIN(x+y/2, y-1); becomes,

=> z = (x+y/2 < y-1)? x+y/2 : y - 1;

=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;

=> z = (3+2 < 4-1)? 3+2 : 4 - 1;

=> z = (5 < 3)? 5 : 3;

The macro return the number 3 and it is stored in the variable z.

Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.

Step 4: printf("%dn", z);. It prints the value of variable z.

Hence the output of the program is 3


3.
What will be the output of the program in TurboC?

#include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10, j=20;
    const int *ptr = &i;
    printf(" i = %5X", ptr);
    printf(" ptr = %d", *ptr);
    ptr = &j;
    printf(" j = %5X", ptr);
    printf(" ptr = %d", *ptr);
    return 0;
}





4.
What will be the output of the program?

#include<stdio.h>

int main()
{
    char huge *near *far *ptr1;
    char near *far *huge *ptr2;
    char far *huge *near *ptr3;
    printf("%d, %d, %dn", sizeof(**ptr1), sizeof(ptr2), sizeof(*ptr3));
    return 0;
}





5.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const int i=0;
    printf("%dn", i++);
    return 0;
}



Explanation:

This program will show an error "Cannot modify a const object".

Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).

Step 2: printf("%dn", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".

Because, we cannot modify a const variable.


6. In which header file is the NULL macro defined?



Explanation:

The macro "NULL" is defined in locale.h, stddef.h, stdio.h, stdlib.h, string.h, time.h, and wchar.h.


7.
Declare the following statement?
"A pointer to an array of three chars".





8.
Which of the following statements are correct about 6 used in the program?
int num[6];
num[6]=21;





9.      A pointer is





10.
What will be the output of the program (in Turbo C)?

#include<stdio.h>

int fun(int *f)
{
    *f = 10;
    return 0;
}
int main()
{
    const int arr[5] = {1, 2, 3, 4, 5};
    printf("Before modification arr[3] = %d", arr[3]);
    fun(&arr[3]);
    printf("nAfter modification arr[3] = %d", arr[3]);
    return 0;
}



Explanation:

Step 1: const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized to

arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5

Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4).

Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10.

A const variable can be indirectly modified by a pointer.

Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10).

Hence the output of the program is

Before modification arr[3] = 4

After modification arr[3] = 10


11.
  What will be the output of the program?

#include

int main()
{
    const c = -11;
    const int d = 34;
    printf("%d, %dn", c, d);
    return 0;
}



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, %dn", c, d); The value of the variable 'c' and 'd' are printed.

Hence the output of the program is -11, 34


12.
What do the following declaration signify?

int *ptr[30];





13.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const char *s = "";
    char str[] = "Hello";
    s = str;
    while(*s)
        printf("%c", *s++);

    return 0;
}



Explanation:

Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.

Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".

Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".

Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.

Hence the output of the program is "Hello".


14.
What do the following declaration signify?

char *arr[10];





15.
    What do the following declaration signify?

char *scr;





16.
What do the following declaration signify?

char **argv;





17.
     What will be the output of the program?

#include
typedef void v;
typedef int i;

int main()
{
    v fun(i, i);
    fun(2, 3);
    return 0;
}
v fun(i a, i b)
{
    i s=2;
    float i;
    printf("%d,", sizeof(i));
    printf(" %d", a*b*s);
}





18.
What will be the output of the program ?

#include<stdio.h>
void fun(int **p);

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0};
    int *ptr;
    ptr = &a[0][0];
    fun(&ptr);
    return 0;
}
void fun(int **p)
{
    printf("%dn", **p);
}





19.
Declare the following statement?
"An array of three pointers to chars".





20.
   What will be the output of the program ?

#include

int main()
{
    char *str;
    str = "%s";
    printf(str, "Kn");
    return 0;
}





21.
What will be the output of the program?

#include<stdio.h>

int main()
{
    int y=128;
    const int x=y;
    printf("%dn", x);
    return 0;
}



Explanation:

Step 1: int y=128; The variable 'y' is declared as an integer type and initialized to value "128".

Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value.

Step 3: printf("%dn", x); It prints the value of variable 'x'.

Hence the output of the program is "128"


22. What will happen if in a C program you assign a value to an array element whose subscript exceeds the size of array?





23.
What will be the output of the program?

#include<stdio.h>

int main()
{
    char far *near *ptr1;
    char far *far *ptr2;
    char far *huge *ptr3;
    printf("%d, %d, %dn", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3));
    return 0;
}





24.
What will be the output of the program ?

#include<stdio.h>

int main()
{
    int a[5] = {5, 1, 15, 20, 25};
    int i, j, m;
    i = ++a[1];
    j = a[1]++;
    m = a[i++];
    printf("%d, %d, %d", i, j, m);
    return 0;
}





25.
What do the following declaration signify?

void (*cmp)();





26.
What will be the output of the program ?

#include<stdio.h>

int main()
{
    int arr[1]={10};
    printf("%dn", 0[arr]);
    return 0;
}





27. In C, if you pass an array as an argument to a function, what actually gets passed?



Explanation:

The statement 'C' is correct. When we pass an array as a funtion argument, the base address of the array will be passed.


28.
What will be the output of the program?

#include<stdio.h>
#define MAN(x, y) ((x)>(y)) ? (x):(y);

int main()
{
    int i=10, j=5, k=0;
    k = MAN(++i, j++);
    printf("%d, %d, %dn", i, j, k);
    return 0;
}



Explanation:

The macro MAN(x, y) ((x)>(y)) ? (x):(y); returns the biggest number of given two numbers.

Step 1: int i=10, j=5, k=0; The variable i, j, k are declared as an integer type and initialized to value 10, 5, 0 respectively.

Step 2: k = MAN(++i, j++); becomes,

=> k = ((++i)>(j++)) ? (++i):(j++);

=> k = ((11)>(5)) ? (12):(6);

=> k = 12

Step 3: printf("%d, %d, %dn", i, j, k); It prints the variable i, j, k.

In the above macro step 2 the variable i value is increemented by 2 and variable j value is increemented by 1.

Hence the output of the program is 12, 6, 12


29.
What will be the output of the program?

#include<stdio.h>
int fun(int **ptr);

int main()
{
    int i=10;
    const int *ptr = &i;
    fun(&ptr);
    return 0;
}
int fun(int **ptr)
{
    int j = 223;
    int *temp = &j;
    printf("Before changing ptr = %5xn", *ptr);
    const *ptr = temp;
    printf("After changing ptr = %5xn", *ptr);
    return 0;
}





30. What would be the equivalent pointer expression for referring the array element a[i][j][k][l]







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