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What will be output of following program?

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#include #include int main(){ int main(){ int a int a = 320; char *ptr; char *ptr; ptr =( char char *)&a; *)&a; printf("%d printf("%d ",*ptr); ",*ptr); return 0; }

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(A) 2 (B) 320

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(C) 64

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(D) Compilation error

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(E) None of above

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Turbo C++ 3.0: 64

pointer

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with

c programming language and

C tricky questions Example of recursion in c

choice

support

far

and

huge

31/07/2013

C pointer s q uesti ons

Merge sort program in c

Turbo C ++4.5: 64 Program to convert decimal to binary in c

Linux GCC: 64 Write a c program to reverse a string Find out the perfect number using c program

Visual C++: 64

Write a c program to find out sum of digit of given number TO FIND FIBONACCI SERIES USING C PROGRAM

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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)

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What will be output of following program?

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#include #include #include #include int main(){ int main(){ void (*p)(); void (*p)(); int (*q)(); int (*q)(); int (*r)(); int (*r)(); p = clrscr;

31/07/2013

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

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

How to test palindrome in c++

return type is int . So they can hold the address of such function.

Palindrome number in c++ Program of palindrome number in c++ C++ code to get sum of all odd numbers in given range

3. What will be output of following program? #include int main(){

31/07/2013

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int i = 3; int *j; int **k; j=&i; k=&j; printf("%u %u %d ",k,*k,**k); return 0;

}

(A) Address, Address, 3 Add to circles

(B) Address, 3, 3 (C) 3, 3, 3 (D) Compilation error (E) None of above Output

Explanation:

Turbo C++ 3.0: Address, Address, 3

Turbo C ++4.5: Address, Address, Address

Linux GCC: Address, Address, 3

Visual C++: Address, Address, 3

Memory representation

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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.

31/07/2013


What will be output of following program? #include int main(){ char far *p =(char far *)0x55550005; char far *q =(char far *)0x53332225; *p = 80; (*p)++; printf("%d",*q); return 0; }

(A) 80 (B) 81 (C) 82 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 81

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error

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

Physical 0x55555

address

of

0x55550005

=

(0x5555)

*

(0x10)

+

(0x0005)

=

31/07/2013


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. What will be output of following program? #include #include int main(){ char *ptr1 = NULL; char *ptr2 = 0; strcpy(ptr1," c"); strcpy(ptr2,"questions"); printf("\n%s %s",ptr1,ptr2); return 0; }

(A) c questions (B) c (null) (C) (null) (null) (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: (null) (null) Turbo C ++4.5: Run time error

31/07/2013


Linux GCC: Run time error Visual C++: Run time error

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

6. What will be output of following program? #include int 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"); return 0; }

(A) power of pointer (B) power of c (C) power of cpower of c (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: power of pointer Turbo C ++4.5: power of c

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Linux GCC: Compilation error Visual C++: Compilation error

Here

we

are

performing

relational

operation

between

two

huge

addresses. So first of all both a and b will normalize as:

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. What will be output of following program? #include #include int main(){ register a = 25; int far *p; p=&a; printf("%d ",*p); return 0; }

(A) 25 (B) 4 (C) Address (D) Compilation error (E) None of above Output

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E x p l a n a t i o n : Turbo C++ 3.0: Compilation error Turbo C ++4.5: Compilation error Linux GCC: Compilation error Visual C++: Compilation error

Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

8. What will be output of following program? #include #include int main(){ char far *p,*q; printf("%d %d",sizeof(p),sizeof(q)); return 0; }

(A) 2 2 (B) 4 4 (C) 4 2 (D) 2 4 (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 4 4

31/07/2013


Turbo C ++4.5: 4 4 Linux GCC: Compilation error Visual C++: Compilation error

p is far pointer which size is 4 byte.

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

9. What will be output of following program? #include int main(){ int a = 10; void *p = &a; int *ptr = p; printf("%u",*ptr); return 0; }

(A) 10 (B) Address (C) 2 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 10

31/07/2013


Turbo C ++4.5: 10 Linux GCC: 10 Visual C++: 10

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

10. What will be output of following program? #include #include int main(){ int register a; scanf("%d",&a); printf("%d",a); return 0; } //if a=25

(A) 25 (B) Address (C) 0 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: Compilation error

31/07/2013





Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

11. What will be output of following program? #include int main(){ char arr[10]; arr = "world"; printf("%s",arr); return 0; }

(A) world (B) w (C) Null (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: Compilation error


31/07/2013




Compilation error Lvalue required

Array name is constant pointer and we cannot assign any value in constant data type after declaration.

12. What will be output of following program? #include #include int main(){ int a,b,c,d; char *p = ( char *)0; int *q = ( int *q)0; float *r = ( float *)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); return 0; }

(A) 2 2 2 2 (B) 1 2 4 8 (C) 1 2 2 4 (D) Compilation error (E) None of above Output

31/07/2013



Turbo C++ 3.0: 1 2 4 8 Turbo C ++4.5: Compilation error Linux GCC: Compilation error Visual C++: Compilation error

Address + 1 = next address

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

next address will be size of data type.

13. What will be output of following program? #include #include int main(){ int a = 5,b = 10,c; int *p = &a,*q = &b; c = p - q; printf("%d" , c); return 0; }

(A) 1 (B) 5 (C) -5

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(D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 1 Turbo C ++4.5: 1 Linux GCC: 1 Visual C++: 2

Difference of two same type of pointer is always one.

14. What will be output of following program? #include unsigned long int (* avg())[3]{ static unsigned long int arr[3] = {1,2,3}; return &arr; } int main(){ unsigned long int (*ptr)[3]; ptr = avg(); printf("%d" , *(*ptr+2)); return 0; }

(A) 1 (B) 2 (C) 3 (D) Compilation error

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(E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 3 Turbo C ++4.5: 3 Linux GCC: 3 Visual C++: 3

15. What will be output of following program? #include int main(){ int * p , b; b = sizeof(p); printf("%d" , b); return 0; }

(A) 2 (B) 4 (C) 8 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 2 or 4

Turbo C ++4.5: 2 or 4

31/07/2013


Linux GCC: 4

Visual C++: 4

since in this question it has not written p is which type of pointer. So

its

output

will

depend

upon

which

Default memory model is small.

16. What will be output of following program? #include int main(){ int i = 5 , j; int *p , *q; p = &i; q = &j; j = 5; printf("%d %d",*p,*q); return 0; }

(A) 5 5 (B) Address Address (C) 5 Address (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 5 5 Turbo C ++4.5: 5 5

memory

model

has

selected.

31/07/2013


Linux GCC: 5 5 Visual C++: 5 5

17. What will be output of following program? #include int main(){ int i = 5; int *p; p = &i; printf(" %u %u", *&p , &*p); return 0; }

(A) 5 Address (B) Address Address (C) Address 5 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: Address Address Turbo C ++4.5: Address Address Linux GCC: Address Address Visual C++: Address Address

Since * and & always cancel to each other.

31/07/2013


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. What will be output of following program? #include int 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); return 0; }

(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) Compilation error (E) None of above

31/07/2013

C pointers questions Output

E x p l a n a t i o n : Turbo C++ 3.0:

value of i : 100 addresss of i : Address


Turbo C ++4.5:



Linux GCC:



Visual C++:

value of i : 100 addresss of i : Address value of i : 101 addresss of i : Address

Within the scope of any variable, value of variable may change but its address will never change in any modification of variable.

19.

31/07/2013


What will be output of following program? #include int main(){ char far *p =(char far *)0x55550005; char far *q =(char far *)0x53332225; *p = 25; (*p)++; printf("%d",*q); return 0; }

(A) 25 (B) Address (C) Garbage (D) Compilation error (E)None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 26 Turbo C ++4.5: Compilation error Linux GCC: Compilation error Visual C++: Compilation error

Far

address

of

p

and

q

are

representing

Physical address of

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

Physical address of

same

physical

address.

31/07/2013


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

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

(*p)++ means to increase the content by one at memory the location 0x5555 so now content of memory location at 0x55555 is 26

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

20. What will be output of following program? #include int main(){ int i = 3; int *j; int **k; j = &i; k = &j; printf("%u %u %u",i,j,k); return 0; }

(A) 3 Address 3 (B) 3 Address Address (C) 3 3 3 (D) Compilation error (E) None of above Output

E x p l a n a t i o n : Turbo C++ 3.0: 3 Address Address

31/07/2013


Turbo C ++4.5: 3 Address Address

Linux GCC: 3 Address Address

Visual C++: 3 Address Address

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

Pointer Tutorial More pointer questions Array questions String questions Function questions +5 Recommend this on Google

12 comments:

31/07/2013

C pointers questions bartek 2/15/12, 2:43 AM

According to Q13, the explanation is wrong. First of all, the answer in this question depends on how this 2 variables are allocated on stack, I mean if they're two consecutive values or they're apart from each other in memory. For example on my Visual C++ in such a declaration: int a = 5, b = 10; Difference between addresses of a and b is 12 bytes, so if they' re 4 bytes ints the difference is &a - &b = 3. Difference between two pointers of same type isn't always 1 !!!!!! It depends on they relative location in memory !!!!!! Reply Replies Lokesh Sharma 6/25/13, 7:55 PM

Difference of two pointers of same type is 1 and not the size of their data types. Try this code on codepad.org which uses linux gcc compiler. Also, the difference between two pointers of same type is not always 1 as mentioned. Try this code #include #include int main(){ int a = 5,b = 10,c = 100, d = -9; int *p = &a,*q = &d; c = p - q; printf("%d" , c); return 0; } Reply

Unknown 8/7/12, 12:12 PM

according to Q14,there is no explanation please provide explanation,it is very good for beginners...otherwise everthing fine... ITS VERY GOOD JOB..PL EASE KEEP IT UP Reply

Laxmipooja Anegun di 8/20/12, 10:55 PM

will this program face dangling pointer problem??? #include unsigned long int (* avg())[3]{ static unsigned long int arr[3] = {1,2,3}; return &arr; }

31/07/2013


int main(){ unsigned long int (*ptr)[3]; ptr = avg(); printf("%d" , *(*ptr+2)); return 0; } Reply

Anonymous 9/14/12, 12:00 AM

i want a C program that takes the input – the mark sheets of N number of students. for examle; ●Enter the marks for student 1 student name: aaa roll number: 24 marks obtained in 5 subjects: 56 89 87 76 98 ●Enter the roll number of the student to be searched: 22 ●The details of student with roll number 22: name: bbb total marks obtained: 293 Reply

Anonymous 9/14/12, 12:03 AM

i need a C program that accepts the roll numbers and names of N students from the user … and prints those in the alphabetical order of the name, along with the roll numbers. Enter the number of students: 3 Enter the name of student #1: vinay Enter vinay’s roll number: 12 Enter the name of student #2: anand Enter anand’s roll number: 14 Enter the name of student #3: kumar Enter kumar’s roll number: 11 The list of students, in sorted order of names is given below: anand (14) kumar (11) vinay (12) Reply

sahil 9/14/12, 12:50 AM

what is O/P of this n please also tell whats the concept behind that o/p? #include main() {

31/07/2013


int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d----%d",*p,*q); } Reply Replies Lokesh Sharma 6/25/13, 7:57 PM

should give an error as q doesn't point to a location. Reply

Neeru Nagi 3/22/13, 5:52 PM

when i put a=908 why i get -116 as output for p rogram 1 Reply


ques no 2 is producing error for pointer r Reply


there is some problem in 9 it is giving error(cant convert void to int) ..?? Reply

Roja 4/22/13, 12:25 AM

7 Question output 25 in C++ 3.0 Reply

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