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Posted Date: 14 May 2008 Resource Type: Articles/Knowledge Sharing Category: Placement Papers
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Posted By: kavinya Member Level: Bronze
Rating:  Points: 1
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section C – analysing program segements
1)struct dlink{
int nodeid;
struct dlink *next;
struct dline *prev;
} dlink_t;
A pointer to the head of that linked list is maintained as a global variable whose definition is
dlink_t *head;
The function remove_element(dlink_t*rp), needs to remove the node pointed to by rp and adjust the head The
first node’s prev and the last node’s text are NULL
remove_element (dlink_t *rp)
{
rp->prev->next =rp->next;
rp->next->prev =rp->prev;
if(head ==rp)
head =rp->next;
} which of the following statement is true about the function remove_element
a)it works when head is the same as rp; b)it does not work whe rp is the last element on the list
c)it sets the head of the list correctly d)it works in all cases (ans.(b))
2.#define NULL 0
char *
index (sp,c)
register char *sp,c;
{
do {
if(*sp==c)
return(sp);
}while (*sp++);
return (NULL); }
The first argument sp,is a pointer to a C string. The second argument c is a character. This function searches
for the character c in the string. If it is found a pointer to that location is returned, else NULL is returned.
This function works
a)Always b)always but fails when the first byte contains the character c
c)works when c is a non NULL character array
d)works only when the character c is found in the string ans. A)Always
3.main() {
printf(“%dn”,f(7)); }
f(x) {
if(x<=4)
return x;
return f(--x); }
a)4 b)5 c)6 d)7 (ans.(4))
4.on a machine where pointers are 4 bytes long,what happens when the following code is executed
main() {
int x=0 ,*p=0;
x++;p++;
printf(“%d and %dn”,p); }
a.1 and 1 is printed b.1 and 4 c.4 and 4 d.causes an exception
5.which is correct?
a)strcpy(char *dst,char *src) {
while (*src)
*dst++=*src++; }
b) strcpy(char *dst,char *src) {
while (*dst++=*src++); }
c) strcpy(char *dst,char *src) {
while (*src){ *dst=*src;
dst++;src++; } }
d) strcpy(char *dst,char *src) {
while (*++dst=*++src); }
6. main() {
int i=20,*j=&i;
f1(j);
*j+=10;
f2(j);
printf(“%d and %d ‘,i,*j); }
f1(k)
int *k;
{ *k+=15;}
f2(x)
int *x;
{ int m=*x, *n=&m;
*n+=10; } The values printed by the program will be
a)20 and 55 b)20 and 45 c)45 and 45 d)55 and 55 e)35 and 35 (ans.(c))or d
7.int
func(int x) {
if(x<=0)
return (1);
return func(x-1)+x; }
main() {
printf(“%d”,func(5)); }
a)12 b)16 c)15 d)11 ans: b
8.consider the following fragments of c code in two files which will be linked together and executed
a.c
int i;
main() {
i=30;
f1();
printf(“%d”,i); }
b.c
static int f1() {
i+=10; } which of the following is true?
a)a.c will fail in compilation phase because f1() is not declared
b)b.c will fail in compilation because the variable i is not declared
c)will print 30 d)a & b
9. void
funca(int *k) {
*k+=20; }
void
funcb(int *k) {
int m=*x,*n=&m;
*n+=10; }
main() {
int var=25,;
*varp=&var;
funca(varp)
*varp+=10;
funcb(varp);
printf("%d%d,var,*varp); }
(a) 20,55(b) 35,35(c) 25,25(d)55,55 (ans. (d))
10. #include
class x{
public :
int a;
x(); };
x::x() { a=10;cout<< a ;}
class b:public x {
public :
b(); x(); };
b::b() { a=20;cout< main() {
b temp; } what will be the output of the following program?
a)10 b)20 c)20 10 d)10 20
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Responses
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| Author: Vidya 22 May 2008 | Member Level: Diamond Points : 2 | gud ariticle
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