a. interaction with the file manager.
b. integrity enforcement.
c. security enforcement.
d. backup and recovery.
e. concurrency control.
1. Mercury
2. Venus
3. Earth
4. Mars
5. Jupiter
<%@ OutputCache Duration="20" Location="Server" VaryByParam="state" VaryByCustom="minorversion" VaryByHeader="Accept-Language"%>
int i=_l_abc(10);
printf("%d\n",--i);
return(i++);
9
- Ÿ Sutkagendar (Baluchistan) – West
- Ÿ Alamgirpur (U.P.) – East
- Ÿ Manda (Jammu) – North
- Ÿ Diamabad (Maharashtra) – South
- Ÿ (Mnemonic – SAMAD)
- Palaeozoic Era
- Mesozoic Era
- Cenozoic Era
int x;
int y;
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
origin is(0,0)
origin is(0,0)
Because DOS stores a file’s seconds in two-second intervals, only the
Page output caching adds the response of page to cache object. Later when page is requested page is displayed from cache rather than creating the page object and displaying it. Page output caching is good if the site is fairly static.
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);
garbagevalue..1
— Gaps between tracks
— Reduce gap to increase capacity
— Same number of bits per track (variable packing density)
— Constant angular velocity
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
30,20,10
— RAID
— Removable
— CD-ROM
— CD-Recordable (CD-R)
— CD-R/W
— DVD
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
1==1 is TRUE
— Do not require all pages of a process in memory
— Bring in pages as required
— Required page is not in memory
— Operating System must swap in required page
— May need to swap out a page to make space
Note :- For this question this post will make a attempt by creating a simple webservice and explaining steps to acheive it. A simple webservice will be created which takes two number and gives addition result of the two number. Definitely the interviewer will not expect such a detail answer but this book will explain you in detail so that you are on right track during interview.
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
2 5 5
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
Compiler error (at line number 4): size of v is Unknown.
public static void main(String [] args)
{
Object a = new Integer(100); // Line1
Object b = new Long(100); // Line2
Object c = new String(“100″); // Line3
a = null; // Line4
a = c; // Line5
c = b; // Line6
b = a; // Line7
// Rest of the code here
}
public static void main(String args[])
{
String s1 = "abc";
String s2 = "abc";
s1 += "xyz";
s2.concat("pqr");
s1.toUpperCase();
System.out.println(s1 + s2);
}
j+=5;
assert(i<5);
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
— In CPU
— May include one or more levels of cache
— “RAM”
Integer i1 = new Integer(50);
Integer i2 = new Integer("50");
Note :- Interviewer will not ask you to name all channels and
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
ck
void *bsearch(const void *key, const void *buf, size_t num, size_t size, int (*comp)(const void *, const void *));
— Nonvolatile
— Very expensive for small runs
String msg = "Hello";
msg += " World";
This is a very important question from practical implementation point of view. Companies who have specific requirement for Remoting projects will expect this question to be answered.
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
g20fy
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
111
222
333
344
Explanation:
Pentium 4 Cache— Data & instructions
For these types of questions interviewer expects small and sweet answers. He is basically looking at what you know about the specific subject. For these type of question this post will provide detail code which is not necessary to be said during interview. Only the basic steps and overall brief are enough to convince that you have knowledge about the subject. Even though this question has detail code and answer say only what is needed in interview.
Above is the figure which shows the three important project
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
100, 100, 100, 2
114, 104, 102, 3
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
Compiler Error
— Fetch instructions
— Interpret instructions
— Fetch data
— Process data
— Write data
CPU Internal Structurechar s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
M
— Add 5 to contents of accumulator
— 5 is operand
class Parent
{
public void print(int i)
{
}
}
public class Child extends Parent
{
public static void main(String argv[])
{
}
// insert code here
}
A. public void print(int i, byte b) throws Exception {}
B. public void print(int i, long i) throws Exception {}
C. public void print(long i) {}
D. public void print(int i) throws Exception {}
E. public int print(int i)
A, B, and C
It is already written.
It is already debugged.
It has been optimized as much as possible (usually).
int (*comp)(const void *ele1, const void *ele2));
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
Compiler error: Lvalue required in function main
