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C Language Questions And Answers Sample Test 2


Sample C Language Test 2 for you to Practice. Evaluate your C Language Test 2 test answering skills by trying the online C Language Sample Test 2 and know your score.

C Language Test 2


You have scored 3 out of 10
You Scored:2
Total Questions:2
Attended Questions:2
Correct Answered:2
1.
What will the SWAP macro in the following program be expanded to on preprocessing? will the code compile?

#include<stdio.h>
#define SWAP(a, b, c)(c t; t=a, a=b, b=t)
int main()
{
    int x=10, y=20;
    SWAP(x, y, int);
    printf("%d %dn", x, y);
    return 0;
}



Explanation:

The code won't compile since declaration of t cannot occur within parenthesis.


2.
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"


3.
    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.


4. 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.


5.
What do the following declaration signify?

int *f();





6.
   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;
}





7.
What will be the output of the program?

#include<stdio.h>

int main()
{
    const int x=5;
    const int *ptrx;
    ptrx = &x;
    *ptrx = 10;
    printf("%dn", x);
    return 0;
}



Explanation:

Step 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'.

Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer.

Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx.

Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error.

To change the value of const variable x we have to use *(int *)&x = 10;


8.
What will be the output of the program in DOS (Compiler - Turbo C)?

#include<stdio.h>
double i;

int main()
{
    (int)(float)(char) i;
    printf("%d", sizeof((int)(float)(char)i));
    return 0;
}



Explanation:

Due to the C language is being platform dependent:

In Turbo C (DOS - 16 bit platform), the output will be 2.

But in GCC (Unix/Linux - 32 bit platform), the output will be 4.


9.
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;
}





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





11.
What will be the output of the program in Turb C (under DOS)?

#include<stdio.h>

int main()
{
    int arr[5], i=0;
    while(i<5)
        arr[i]=++i;

    for(i=0; i<5; i++)
        printf("%d, ", arr[i]);

    return 0;
}





12.
     What will be the output of the program?

#include
#define MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c)

int main()
{
    int x;
    x = MAX(3+2, 2+7, 3+7);
    printf("%dn", x);
    return 0;
}



Explanation:

The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

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

Hence the output of the program is "10".


13.
     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);
}





14.
What will be the output of the program ?

#include<stdio.h>

int main()
{
    static char *s[] = {"black", "white", "pink", "violet"};
    char **ptr[] = {s+3, s+2, s+1, s}, ***p;
    p = ptr;
    ++p;
    printf("%s", **p+1);
    return 0;
}





15.
Declare the following statement?
"An array of three pointers to chars".





16.    What is (void*)0?





17.
Declare the following statement?
"A pointer to an array of three chars".





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

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





20. The operator used to get value at address stored in a pointer variable is





21. 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.


22.
What will be the output of the program if the array begins 1200 in memory?

#include<stdio.h>

int main()
{
    int arr[]={2, 3, 4, 1, 6};
    printf("%u, %u, %un", arr, &arr[0], &arr);
    return 0;
}





23.
Declare the following statement?
"A pointer to a function which receives an int pointer and returns float pointer".





24. 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.


25.
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);
}





26.
What do the following declaration signify?

void (*cmp)();





27.
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.


28. What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?
#include<stdio.h>

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
    printf("%u, %un", a+1, &a+1);
    return 0;
}





29.
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;
}





30.
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






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