C Puzzle Answers

Answers L1.A1 *Solution for L1.Q1 a = xxx * 10 which is => a => a => a => a

= = = =

ABC - XYZ * 10 20 - 10 * 10 20 - 100 -80

L1.A2 Solution for L1.Q2 Actual substitution is like this : calc(20+4, 10 -2) is calculated as follows

since

(20+4 * 10-2) / (20+4 - 10-2) (20+40-2) / 12 58 / 12 = 4.8 it is printed in %d the ans is 4

L1.A3 Solution for L1.Q3 This problem will compile properly, but it will give run time error. It will give divide-by-zero error. Look in to the do loop portion do { a /= cnt; } while (cnt --); when the 'cnt' value is 1, it is decremented in 'while ( cnt --)' and on next reference of 'cnt' it becomes zero. a /= cnt; /* ie. a /= 0 */ which leads to divide-by-zero error.

L1.A4 Solution for L1.Q4 the result will be c = 40 and abc = 0; because the scope of the variable 'abc' inside if(c) {.. is not valid out side that if (.) { .. }.

}

L1.A5 Solution for L1.Q5 The answer is 7.

The first condition ++k < 5 is checked and

it is false (Now k = 6). So, it checks the 3rd condition (or condition ++k <= 8) and (now k = 7) it is true. At this point k value is incremented by twice, hence the value of k becomes 7.

L1.A6 Solution for L1.Q6 The solution depends on the implementation of stack. (Depends on OS) In some machines the arguments are passed from left to right to the stack. In this case the result will be Main :

5 7 Fn :

7 7

Other machines the arguments may be passed from right to left to the stack. In that case the result will be Main : 6 6 Fn : 8 7

L2.A1 Solution for L2.Q1 The output will be : 0 10 11 20 21 *s++ *++s ++*s

=> *(s++) => *(++s) => ++(*s)

L2.A2 Solution for L2.Q2 Some compiler (ansi) may give warning message, but it will compile without errors. The output will be : 0 1 2 3 4 and -1

L2.A3 Solution for L2.Q3 This is the problem about Unions. Unions are similar to structures but it differs in some ways. Unions can be assigned only with one field at any time. In this case, unions x and y can be assigned with any of the one field a or b or c at one time. During initialisation of unions it takes the value (whatever assigned ) only for the first field. So, The statement y = {100} intialises the union y with field a = 100. In this example, all fields of union x are assigned with some values. But at any time only one of the union field can be assigned. So, for the union x the field c is assigned as 21.50. Thus, The output will be Union 2 : 22 22 21.50 Union Y : 100 22 22 ( 22 refers unpredictable results )

L2.A4 Solution for L2.Q4 The pointer x points to the same location where y is stored. So, The changes in y reflects in x. The output will be : 10 11 30 31

20 21

40 41

L2.A5 Solution for L2.Q5 The

output

of

this

program

is

purely

depends

on

the

processor architecuture. If the sizeof integer is 4 bytes and the size of character is 1 byte (In some computers), the output will be sizeof xyz = 20

sizeof abc = 100

The output can be generalized to some extent as follows, sizeof xyz = 4 * sizeof(int) + 1 * sizeof(char) + padding bytes sizeof abc = 100 * sizeof(char) + padding bytes To keep the structures/unions byte aligned, some padding bytes are added in between the sturcture fields. In this example 3 bytes are padded between ' char ch' and 'int d' fields. The unused bytes are called holes. To understand more about padding bytes (holes) try varing the field types of the structures and see the output.

L2.A6 Solution for L2.Q6 Just try to execute this file as such. You can find out that it will exit immediately. Do you know why? With this hint, we can trace out the error. If you look into the macro 'Error', you can easily identify that there are two separete statements without brases '{ ..}'. That is the problem. So, it exits after the calling open(). The macro should be put inside the brases like this. #define Error(str)

{ printf("Error : %s\n", str); exit(1); }

L2.A7 Solution for L2.Q7 This program will fault (Memory Can you predict Why?

fault/segmentation

fault).

Remove the statment 'free(a);' from the program, then execute the program. It will run. It gives the results correctly. What causes 'free(a)' to generate fault? Just trace the address location of pointer variable 'a'. The variable 'a' is incremented inside the 'for loop'. Out side the 'for loop' the variable 'a' will point to 'null'. When the free() call is made, it will free the data area from the base_address (which is passed as the argument of the free call) upto the length of the data allocated previously. In this case, free() tries to free the length of 10 *sizeof(int) from the base pointer location passed as the argument to the free call, which is 'null' in this case.

Thus, it generates memory fault.

L2.A8 Solution for L2.Q8 #include main(argc, argv) int argc; char *argv[]; { int count = 0, i; int v = atoi(argv[1]); for(i=0; i<8*sizeof(int); i++) if(v &(1<< } count); v, %d='%d\n",' in 1?s of printf(?No count++;>

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 func1() and func2() then again func2().

LX.A8 Solution for LX.Q8 #include

called :

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 int

*new; 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 output as follows Hello! how is this? That is C

problem.

super

You will

get the