C Puzzles : Questions and Answers - Level 3

Questions

L3.Q1 :

 Write a function revstr() - which reverses the given string
 in the same string buffer using  pointers.  (ie) Should not
 use extra buffers for copying the reverse string.

Solution for L3.Q1

L3.Q2 :

Write a program  to print the series 2 power x, where x >= 0
( 1, 2, 4, 8, 16, ....  ) without  using C math  library and
arithmatic  operators  ( ie.  *, /, +, - and  math.h are not
allowed)

Solution for L3.Q2

L3.Q3 :

Write a  program  to swap two  integers  without  using  3rd
integer (ie.  Without using any temporary variable)

Solution for L3.Q3

L3.Q4 :

Write a general swap macro in C :

  - A macro which can swap any type of data (ie.  int, char,
	float, struct, etc..)

Solution for L3.Q4

L3.Q5 :

Write a program to delete  the entry from the doubly  linked
list  without  saving any of the  entries of the list to the
temporary variable.

Solution for L3.Q5

L3.Q6 : What will be the output of this program ?

#include 

main()
{
	int	*a, *savea, i;

	savea = a = (int *) malloc(4 * sizeof(int));

	for (i=0; i<4; i++) *a++ = 10 * i;


	for (i=0; i<4; i++) {
		printf("%d\n", *savea);
		savea += sizeof(int);
	}
}

Solution for L3.Q6

LX.Q7 : Trace the program and print the output

#include 

typedef	int abc(int a, char *b);


int func2(int a, char *b)
{
	a *= 2;
	strcat(b, "func2 ");
	return a;
}

int func1(int a, char *b)
{
	abc	*fn = func2;

	a *= a;
	strcat(b, "func1 ");
	return (fn(a, b));
}

main()
{
	abc	*f1, *f2;
	int 	res; 
	static  char	str[50] = "hello! ";

	f1 = func1;
	res = f1(10, str);
	f1 = func2;
	res = f1(res, str);

	printf("res : %d  str :  %s\n", res, str);
}

Solution for LX.Q7

LX.Q8 :

Write a program to reverse a Linked list within the same list 

Solution for LX.Q8

LX.Q9 : What will be the output of this program

#include 

main()
{
  int a=3, b = 5;

  printf(&a["Ya!Hello! how is this? %s\n"], &b["junk/super"]);
  printf(&a["WHAT%c%c%c  %c%c  %c !\n"], 1["this"],
	 2["beauty"],0["tool"],0["is"],3["sensitive"],4["CCCCCC"]);
}

Solution for LX.Q9

LX.Q10 :

Solution for LX.Q10

Answers

L3.A1
Solution for L3.Q1

#include 

char *rev_str(char *str)
{
	char *s = str, *e = s + strlen(s) -1;
	char *t = "junk"; /* to be safe - conforming with ANSI C std */

	while (s < e) { *t = *e; *e-- = *s; *s++ = *t; } 
	return(str);
}

/* Another way of doing this */

char *str_rev(char *str)
{

	int len = strlen(str),i=0,j=len/2;
	
	len--;
	while(i < j) {
		*(str+i)^=*(str+len)^=*(str+i)^=*(str+len);
		i++; len--;
	}
	return(str);
}

main (int argc, char **argv)
{
	printf("1st method : %s\n", rev_str(argv[1]));
	printf("2nd method : %s\n", str_rev(argv[1]));
}

L3.A2
Solution for L3.Q2

#include 

void main()
{
	int i;

	for(i=0; i< 10; i++)
		printf("%d\t", 2 << i);
}

L3.A3
Solution for L3.Q3

#include 

main()
{
	int a, b;

	printf("Enter two numbers A, B : ");
	scanf("%d %d", &a, &b);

	a^=b^=a^=b;	/* swap A and B */

	printf("\nA = %d, B= %d\n", a, b);
}

L3.A4
Solution for L3.Q4

#include 

/* Generic Swap macro*/
#define swap(a, b, type) { type t = a; a = b; b = t; }

/* Verification routines */
main()
{
	int a=10, b =20;
	float e=10.0, f = 20.0;
	char *x = "string1", *y = "string2";
	typedef struct { int a; char s[20]; } st;
	st	s1 = {50, "struct1"}, s2 = {100, "struct2"};

	swap(a, b, int);
	printf("%d %d\n", a, b);

	swap(e, f, float );
	printf("%f %f\n", e, f);

	swap(x, y, char *);
	printf("%s %s\n", x, y);

	swap(s1, s2, st);
	printf("S1: %d %s \tS2: %d %s\n", s1.a, s1.s, s2.a, s2.s); 

	ptr_swap();
}

ptr_swap()
{
	int *a, *b;
	float *c, *d;

	a = (int *) malloc(sizeof(int));
	b = (int *) malloc(sizeof(int));
	*a = 10; *b = 20;

	swap(a, b, int *);
	printf("%d %d\n", *a, *b);

	c = (float *) malloc(sizeof(float));
	d = (float *) malloc(sizeof(float));
	*c = 10.01; *d  = 20.02;

	swap(c, d, float *);
	printf("%f %f\n", *c, *d);
}

L3.A5
Solution for L3.Q5

#include 

/* Solution */

typedef struct Link{
	int  	    val;
	struct Link *next;
	struct Link *prev;
} Link;

void DL_delete(Link **, int);

void DL_delete(Link **head, int val)
{
	Link	**tail;

	while ((*head)) {
		if ((*head)->next == NULL) tail = head;
		if ((*head)->val == val) {
			*head = (*head)->next;
		}
		else head = &(*head)->next;
	}
	while((*tail)) {
		if ((*tail)->val == val) {
			*tail = (*tail)->prev;
		}
		else tail= &(*tail)->prev;
	}
}

/* Supporting (Verification) routine */

Link *DL_build();
void DL_print(Link *);

main()
{
	int	val;
	Link	*head;

	head = DL_build();
	DL_print(head);

	printf("Enter the value to be deleted from the list : ");
	scanf("%d", &val);

	DL_delete(&head, val);
	DL_print(head);
}

Link *DL_build()
{
	int 	val;
	Link	*head, *prev, *next;

 	head = prev = next =  NULL;

	while(1) {
		Link *new;

		printf("Enter the value for the list element (0 for end) : ");
		scanf("%d", &val);

		if (val == 0) break;

		new = (Link *) malloc(sizeof(Link));
		new->val = val;
		new->prev = prev;
		new->next = next;

		if (prev) prev->next = new;
		else head = new;

		prev = new;
	}

	return (head);
}

void DL_print(Link *head)
{
	Link *shead = head, *rhead;
	printf("\n****** Link List values ********\n\n");
	while(head) {
		printf("%d\t", head->val);
		if (head->next == NULL) rhead = head;
		head = head->next;
	}

	printf("\n Reverse list \n");
	while(rhead)
	{
		printf("%d\t", rhead->val);
		rhead = rhead->prev;
	}
	printf("\n\n");
}

L3.A6
Solution for L3.Q6

The first value will be 0, the rest of the three values will
not be  predictable.  Actually  it prints the  values of the
following location in each step

	* savea
	* (savea + sizeof(int) * sizeof(int))
	 etc...

ie. savea += sizeof(int) => savea = savea + 
				sizeof(savea_type) * sizeof(int)
			  ( by pointer arithmatic)
			 => save = savea +  sizeof(int) * sizeof(int)

Note: You can verify the above by varing the type of 'savea'
	  variable to char, double, struct, etc.

Instead of statement 'savea += sizeof(int)' use savea++ then
the values 0, 10, 20 and 30 will be printed.  This behaviour
is because of pointer arithmatic.

LX.A7
Solution for LX.Q7

Two  function  pointers f1 and f2 are  declared  of the type
abc.  whereas abc is a pointer to a function returns int.

func1()  which  is  assigned  to  f1  is  called  first.  It
modifies  the  values of the  parameter  'a' and 'b' to 100,
"hello!  func1  ".  In  func1(),  func2()  is  called  which
further  modifies  the value of 'a' and 'b' to 200,  "hello!
func1  func2 " and  returns the value of 'a' which is 200 to
the main.  Main calls f1() again after assigning  func2() to
f1.  So,  func2() is called  and it  returns  the  following
value which will be the output of this program.

	res : 400	str : hello! func1 func2 func2

The output string shows the trace of the functions  called :
func1() and func2() then again func2().

LX.A8
Solution for LX.Q8

#include 

typedef struct  Link {
	int  val;
	struct Link *next;
} Link;

/* Reverse List function */
Link *SL_reverse(Link *head)
{
	Link 	 *revlist =  (Link *)0;

	while(head) {
		Link *tmp;

		tmp = head;
		head = head->next;
		tmp->next = revlist;
		revlist = tmp;
	}

	return revlist;
}

/* Supporting (Verification) routines */

Link *SL_build();

main()
{
	Link	*head;

	head = SL_build();
	head = SL_reverse(head);

	printf("\nReversed List\n\n");
	while(head) { 
		printf("%d\t", head->val);
		head = head->next;
	}
}


Link *SL_build()
{
	Link *head, *prev;

	head = prev = (Link *)0;

	while(1) {
		Link	*new;
		int 	val;

		printf("Enter List element [ 0 for end ] : ");
		scanf("%d", &val);
		if (val == 0) break;

		new = (Link *) malloc(sizeof(Link));
		new->val = val;
		if (prev) prev->next = new;
		else head = new;
		prev = new; 
	}

	prev->next = (Link *)0;
	return head;
}

LX.A9
Solution for LX.Q9

In C we can index an array in two ways. 
For example look in to the following  lines
	int a[3] = {10, 20, 30, 40};
	In this example  index=3 of array 'a' can be represented
	in 2 ways.
	1) a[3] and   
	2) 3[a]
i.e) a[3] = 3[a] = 40

Extend  the same  logic to this  problem.  You will  get the
output as follows

	Hello! how is this?  super
	That is C

LX.A10
Solution for LX.Q10