c pointer objective questions with answers

1) What will be output of following program?

void main()

{

int a=320;

char *ptr;

ptr=(char *)&a;

printf("%d ",*ptr);

getch();

}

(a) 2

(b) 320

(c) 64

(d)Compiler error

(e)None of above

 (2) What will be output of following program?

#include"stdio.h"

#include"conio.h"

void main()

{

void (*p)();

int (*q)();

int (*r)();

p=clrscr;

q=getch;

r=puts;

(*p)();

(*r)("cquestionbank.blogspot.com");

(*q)();

}

(a) NULL

(b) cquestionbank.blogspot.com

(c) c

(d) Compiler error

(e) None of above

(3) What will be output of following program?

void main()

{

int i=3;

int *j;

int **k;

j=&i;

k=&j;

printf(“%u %u %d ”,k,*k,**k);

}

(a) Address, Address, 3

(b)Address, 3, 3

(c) 3, 3, 3

(d) Compiler error

(e) None of above

 (4) What will be output of following program?

void main()

{

char far *p=(char far *)0x55550005;

char far *q=(char far *)0x53332225;

*p=80;

(*p)++;

printf("%d",*q);

getch();

}

(a) 80

(b) 81

(c) 82

(d) Compiler error

(e) None of above

 (5) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

char *ptr1=NULL;

char *ptr2=0;

strcpy(ptr1," c");

strcpy(ptr2,"questions");

printf("\n%s %s",ptr1,ptr2);

getch();

}

(a) c questions

(b) c (null)

(c) (null) (null)

(d) Compiler error

(e) None of above

 (6) What will be output of following program?

void main()

{

int huge *a=(int huge *)0x59990005;

int huge *b=(int huge *)0x59980015;

if(a==b)

printf("power of pointer");

else

printf("power of c");

getch();

}

(a) power of pointer

(b) power of c

(c) power of cpower of c

(d) Compiler error

(e) None of above

 (7) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

register a=25;

int far *p;

p=&a;

printf("%d ",*p);

getch();

}

(a) 25

(b) 4

(c) Address

(d) Compiler error

(e) None of above

 (8) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

char far *p,*q;

printf("%d %d",sizeof(p),sizeof(q));

getch();

}

(a) 2 2

(b) 4 4

(c) 4 2

(d) 2 4

(e) None of above

 (9) What will be output of following program?

void main()

{

int a=10;

void *p=&a;

int *ptr=p;

printf("%u",*ptr);

getch();

}

(a) 10

(b) Address

(c) 2

(d) Compiler error

(e) None of above

 (10) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

int register a;

scanf("%d",&a);

printf("%d",a);

getch();

}

//if a=25

(a) 25

(b) Address

(c) 0

(d) Compiler error

(e) None of above

 (11) What will be output of following program?

void main()

{

char arr[10];

arr="world";

printf("%s",arr);

getch();

}

(a) world

(b) w

(c) Null

(d) Compiler error

(e) None of above

 (12) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

int a,b,c,d;

char *p=0;

int *q=0;

float *r=0;

double *s=0;

a=(int)(p+1);

b=(int)(q+1);

c=(int)(r+1);

d=(int)(s+1);

printf("%d %d %d %d",a,b,c,d);

}

(a)2 2 2 2

(b)1 2 4 8

(c)1 2 2 4

(d) Compiler error

(e) None of above

 (13) What will be output of following program?

#include"stdio.h"

#include"string.h"

void main()

{

int a=5,b=10,c;

int *p=&a,*q=&b;

c=p-q;

printf("%d",c);

getch();

}

(a) 1

(b) 5

(c) -5

(d) Compiler error

(e) None of above

(14) What will be output of following program?

unsigned long int (* avg())[3]

{

static unsigned long int arr[3]={1,2,3};

return &arr;

}

void main()

{

unsigned long int (*ptr)[3];

ptr=avg();

printf("%d",*(*ptr+2));

getch();

}

(a) 1

(b) 2

(c) 3

(d) Compiler error

(e) None of above

 (15) What will be output of following program?

void main()

{

int * p,b;

b=sizeof(p);

printf(“%d”,b);

}

(a) 2

(b) 4

(c) 8

(d) Compiler error

(e) None of above

(16) What will be output of following program?

void main()

{

int i=5,j;

int *p,*q;

p=&i;

q=&j;

j=5;

printf("value of i : %d value of j : %d",*p,*q);

getch();

}

(a) 5 5

(b) Address Address

(c) 5 Address

(d) Compiler error

(e) None of above

(17) What will be output of following program?

{

int i=5;

int *p;

p=&i;

printf(" %u %u",*&p,&*p);

getch();

}

(a) 5 Address

(b) Address Address

(c) Address 5

(d) Compiler error

(e) None of above

(18) What will be output of following program?

void main()

{

int i=100;

printf("value of i : %d addresss of i : %u",i,&i);

i++;

printf("\nvalue of i : %d addresss of i : %u",i,&i);

getch();

}

(a) value of i : 100 addresss of i : Address

value of i : 101 addresss of i : Address

(b) value of i : 100 addresss of i : Address

value of i : 100 addresss of i : Address

(c) value of i : 101 addresss of i : Address

value of i : 101 addresss of i : Address

(d) Compiler error

(e) None of above

 (19) What will be output of following program?

void main()

{

char far *p=(char far *)0x55550005;

char far *q=(char far *)0x53332225;

*p=25;

(*p)++;

printf("%d",*q);

getch();

}

(a) 25

(b) Address

(c) Garbage

(d) Compiler error

(e) None of above

(20) What will be output of following program?

void main()

{

int i=3;

int *j;

int **k;

j=&i;

k=&j;

printf(“%u %u %u”,i,j,k);

}

(a) 3 Address 3

(b) 3 Address Address

(c) 3 3 3

(d) Compiler error

(e) None of above

Answer:

(1) c

(2) b

(3) a

(4) b

(5) c

(6) a

(7) d

(8) c

(9) a

(10) d

(11) d

(12) b

(13) a

(14) c

(15) e

(16) a

(17) b

(18) a

(19) e

(20) b

Explanation

(1)

As we know int is two byte data byte while char is one byte data byte. char pointer can keep the address one byte at time.

Binary value of 320 is 00000001 01000000 (In 16 bit)

Memory representation of int a=320 is:

So ptr is pointing only first 8 bit which color is green and

Decimal value is 64.

(2)

p is pointer to function whose parameter is void and return type

is also void. r and q is pointer to function whose parameter is

void and return type is int . So they can hold the address of

such function.

(3)

Memory representation

Here 6024, 8085, 9091 is any arbitrary address, it may be different.

Value of k is content of k in memory which is 8085

Value of *k means content of memory location which address k keeps.

k keeps address 8085 .

Content of at memory location 8085 is 6024

In the same way **k will equal to 3.

Short cut way to calculate:

Rule: * and & always cancel to each other

i.e. *&a=a

So *k=*(&j) since k=&j

*&j=j =6024

And

**k=**(&j)=*(*&j)=*j=*(&i)=*&i=i=3

(4)

Far address of p and q are representing same physical address .

Physical address of 0x55550005= (0x5555)*(0x10) + (0x0005) = 0x55555

Physical address of 0x53332225= (0x5333*0x10) + (0x2225) =0x55555

*p =80, means content at memory location 0x55555 is assigning value 25

(*p)++ means increase the content by one at memory location 0x5555 so now

content at memory location 0x55555 is 81

*q also means content at memory location 0x55555 which is 26

(5)

We cannot assign any string constant in null pointer by strcpy function.

(6)

Here we are performing relational operation between two huge addresses. So

first both a and b will normalize.

a= (0x5999)* (0x10) + (0x0005) =0x9990+0x0005=0x9995

b= (0x5998)* (0x10) + (0x0015) =0x9980+0x0015=0x9995

Here both huge addresses are representing same physical address. So a==b

is true.

(7)

Register data type stores in CPU. So it has not any memory

address. Hence we cannot write &a.

(8)

p is far pointer which size is 4 byte.

By default q is near pointer which size is 2 byte.

(9)

void pointer can hold address of any data type without type casting. Any

pointer can hold void pointer without type casting.

(10)

Register data type stores in CPU. So it has not any memory address. Hence

we cannot write &a.

(11) Compiler error Lvalue required

Array name is constant pointer and we cannot assign any

value in constant data type after declaration.

(12)

Address=next address

Since initial address of all data type is zero. So its

next address will be size of data type.

(13)

Difference of two same type of pointer is always one.

(15)

Output: 2 or 4

since in this question it has not written p is which type

of pointer. So its output will depend upon which memory

model has selected. Default memory model is small.

(17)

Since * and & always cancel to each other.

i.e. *&a=a

so *&p=p which store address of integer i

&*p=&*(&i) //since p=&i

=&(*&i)

=&i

So second output is also address of i

(18)

Within the scope of any variable, value of variable

may change but its address will never change in any

modification of variable.

(19)

Far address of p and q are representing same physical

address. Physical address of

0x55550005= 0x5555*ox10+ox0005= 0x55555

Physical address of

0x53332225=0x5333*0x10+ox2225=0x55555

*p =25, means content at memory location 0x55555 is

assigning value 25

(*p)++ means increase the content by one at memory

location 0x5555 so now content at memory location 0x55555

is 26

*q also means content at memory location 0x55555 which is

26

(20)

Here 6024, 8085, 9091 is any arbitrary address, it may be different.

Read more: http://cmagical.blogspot.com/2009/11/pointer-objective-questions-with.html#ixzz0y172akrn
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Tricky questions on C

1)Write a C program to find a peculiar two digit number which is three times the sum of its digits.

2) Bacteria are known to multiply very rapidly. If a certain container contains just one bacterium on the first day and there are twice as many on the next day. In this manner the number of bacteria in the container doubles itself everyday. Assuming that the container would be full on the 10th day with 13,312 bacteria, find the number of bacteria that was initially in the container on the first day.

3) Calculate the factorial of a number recursively. From that calculate the value of COS(X) = 1 - X2/2! + X4 /4! - X6/ 6! +…….

4) A number of “Cats” got together and decided to kill between them 999919 mice. Every cat killed equal number of “mice”. Write a program to find number of cats.

5) Consider the following number 45*45=2025; 20+25=45.Write a program to generate number between 32 and 99 that satisfies the above property.

6) Rita has a money pouch containing Rs.700. There are equal number of 25 paise coins, 50 paise and one rupee coins. Write a C program to find how many of each are there?

7) Calculate the factorial of a number recursively and from that calculate the value of ex=1+ (X1/1!) + (X2 /2!) + (X3/3!) + ……….

8) There are some goats and ducks in a farm. There are 60 eyes and 86 foot in total. Write a program to find number of goats and ducks in the farm.

9) Write a C program to find a three digit number which is greater than the aggregate of its third, tenth and the twelfth parts by 58.

10) Write a C program to find a two digit number, the second digit of which is smaller than its first digit by 4, and if the number was divided by the digit’s sum, the quotient would be 7.

Read more: http://cmagical.blogspot.com/2009/11/technical-question-31-of-32-print.html#ixzz0y16ofSCW

datatypes Questions on c

(1) How many type of data c?
Ans:
In c, there are three types of data.

(2) What is qualifier or modifier of data type?
Ans:
It qualify the primary data type.There are five group of qualifier in C.

(There is another type qualifier near, far, huge, which qualify only pointer type data type ,interrupt is also qualifier of data)
We can write all five (one for each group) qualifiers for same data type. If we will not write then it will take its default quantifier.We can write quantifier in any order.We cannot write two qualifier of the same group.
E.g short unsigned volatile const int a=5;
It is right. Qualifier of each group is:

E.g static auto int a=5;
It is wrong, because we cannot write two qualifier of the same group.Both static and auto comes under group 1 i.e storage class.
Note. There is not any keyword for the opposite const and volatile .So if does not write qualifier const it means the default qualifier is opposite of const.Same this happen for volatile.Default qualifier of group 3 has also not any keyword .It is between the short and long i.e normal data type.
It not necessary that each primary data type support all five group of quantifier.

g

(3) What is size of each data type?

Ans:
Size of data type depends upon microprocessor.
Size of int is word length
Size of short int can be >= word length/2 but <=word length Size of long int can be <= 2*word length but >=word length
Size of char, float, double, long double is always fix.
Size of enum is size of int.
Turbo C is based on 8086 microprocessor and its word length is two byte

For TURBO C
Data type size (in byte)
char 1
Short int 2
int 2
long int or long 4
enum 2
float 4
double 8
long double 10
(4) What is const and volatile qualifier?
Ans:
Value of any variable can be changed either by program or external device.const. and volatile is not opposite to each other.
Const : when any variable has qualified with const keyword in declaration statement then later it is not possible to assigne any value or modify by the program.But indirectly with the help of pointer its value can be changed.
When any variable is not qualified by const variable then it’s default qualifier is not const .There is not any keyword for not const.It’s meaning is that value of variable can be changed after the declaration statement by the program.
e.g
What will be output?
void main()
{
const int a=5;
a++;


printf(“%d”,a);
}
Ans: compiler error, we cannot modify const variable.

Volatile: when any variable has qualified by volatile keyword in declaration statement then value of variable can be changed by any external device or hardware interrupt.
If any variable has not qualified by volatile keyword in declaration statement, then then compiler will take not volatile as default quantifier .There is not any special keyword for not volatile.Not volatile means when any variable has qualified by volatile keyword in declaration statement then value of variable cannot be changed by any external device or hardware interrupt.
(5) What is meaning of the declaration:
const volatile int a=6;
Ans:
Value of variable cannot be changed by program (due to const) but its value can be changed by external device or hardware interrupt (due to volatile).

(9) How char data type is represented in the memory?
Ans:
char data type may be signed or unsigned .Both has different memory representation.Both are 8 bit data type.
unsigned char:
All 8 bit is used as data bit.
e.g memory representation of unsigned char a= 7;
Binary equivalent of 7 is 111
For 8 bit we will add 5 zero in the left side i.e 00000111
In the memory:

Here MSD is most significant digit and LSD is list significant digit.
signed char:
1 bit: signed bit
7 bit: data bit
Note: In C, negative number is stored in the 2’s complement format.
Signed bit is 0: Number is positive.
Signed bit is 1: Number is negative.

e.g memory representation of char a=7;
Binary equivalent of 7 is 111
For 8 bit we will add 5 zero in the left side i.e 00000111
Memory representation:

e.g memory representation of char a= -7;
Binary equivalent of 7 is 111
For 8 bit we will add 5 zero in the left side i.e 00000111
Since a is negative number so it will store in the memory in the 2’s complement format

Memory representation:
.

(10) What is endianness of processor?
Ans:
If the size of data type is more than one byte then endianness decides the memory representation of data type.There are two type of endianness.
Little-endian: The processor which follow the following memory representation of data is known as little-endian processor.


First A will fill then B then C then D then E and so on from right to left.
Example of processor: 8085,8086,8088,80286,80386,p1,p2 etc.
Big-endian: The processor which follow the following memory representation of data is known as big-endian processor.

First A will fill then B then C then D then E from right to left.
Example of processor: Motorola, IBM PC etc.