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C Language Questions And Answers Sample Test 9


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C Language Test 9


You have scored 3 out of 10
You Scored:2
Total Questions:2
Attended Questions:2
Correct Answered:2
1. The operator used to get value at address stored in a pointer variable is





2.
What does the following declaration mean?
int (*ptr)[10];





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



Explanation:

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


4.
    What do the following declaration signify?

char *scr;





5.
What will be the output of the program?

#include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int va1=10;
    int va12=20;
    printf("%dn", FUN(va1, 2));
    return 0;
}



Explanation:

The following program will make you understand about ## (macro concatenation) operator clearly.


6.
Declare the following statement?
"An array of three pointers to chars".





7.
What do the following declaration signify?

int (*pf)();





8.
What do the following declaration signify?

int *ptr[30];





9.
Which of the following statements are correct about 6 used in the program?
int num[6];
num[6]=21;





10.
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"


11.
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;
}





12.
What do the following declaration signify?

void (*cmp)();





13.
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.





14.
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;
}





15.
      What will be the output of the program ?

#include

int main()
{
    int i=3, *j, k;
    j = &i;
    printf("%dn", i**j*i+*j);
    return 0;
}





16. If a variable is a pointer to a structure, then which of the following operator is used to access data members of the structure through the pointer variable?





17.
What do the following declaration signify?

char *arr[10];





18.
In which stage the following code 
#include<stdio.h> 
gets replaced by the contents of the file stdio.h



Explanation:

The preprocessor replaces the line #include <stdio.h> with the system header file of that name. More precisely, the entire text of the file 'stdio.h' replaces the #include directive.


19.
What will be the output of the program?

#include<stdio.h>
#include<stdlib.h>

union employee
{
    char name[15];
    int age;
    float salary;
};
const union employee e1;

int main()
{
    strcpy(e1.name, "K");
    printf("%s %d %f", e1.name, e1.age, e1.salary);
    return 0;
}



Explanation:

The output will be (in 16-bit platform DOS):

K 75 0.000000


20.
What do the following declaration signify?

char **argv;





21.
Declare the following statement?
"A pointer to an array of three chars".





22.
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;
}





23.
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.


24.
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".


25.
What will be the output of the program ?

#include<stdio.h>

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





26.
     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);
}





27.
   What will be the output of the program ?

#include

int main()
{
    float arr[] = {12.4, 2.3, 4.5, 6.7};
    printf("%dn", sizeof(arr)/sizeof(arr[0]));
    return 0;
}





28.
What will be the output of the program?

#include<stdio.h>
int get();

int main()
{
    const int x = get();
    printf("%d", x);
    return 0;
}
int get()
{
    return 20;
}



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".


29.
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


30.
What do the following declaration signify?

int *f();







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