7/24/08

C Puzzles: Questions and Answers


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.

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

arithmetic operators ( ie. *, /, +, - and math.h are not

allowed)

L3.Q3 :

Write a program to swap two integers without using 3rd

integer (ie. Without using any temporary variable)

L3.Q4 :

Write a general swap macro in C :

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

float, struct, etc..)

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.

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;>

for (i=0; i<4;>

printf("%d\n", *savea);

savea += sizeof(int);

}

}

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);

}

LX.Q8 :

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

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"]);

}

LX.Q10 : Answers

L3.A1
Slution 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 < t =" *e;">

return(str);

}

/* Another way of doing this */

char *str_rev(char *str)

{

int len = strlen(str),i=0,j=len/2;

len--;

while(i <>

*(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<>

printf("%d\t", 2 <<>

}

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

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