t=0;
for(i=n; i>0; i--)
{
  t|=(1<<p);
  p++;
}
t=t&y;
x=x&t;

Sorry above answer ; I did a mistake
t=0;
for(i=n; i>0; i--)
{
  t|=(1<<p);
  p++;
}
x=x&~t
t=t&y;
x=x|t;

t =((~(~0<<n))<<p)
x = x & ~t;
y = y & t;
x=x|y;

It should be p--, not p++. for example y = 217 (11011001) 
and you want to extract the least 4 SB then n = 4 and p = 3 
(remember the number start from 0 to 7 in memory). After 
executing this you will get x = 9 i.e 1001.

t=0;
for(i=n; i>0; i--)
{
      t |= (1<<p);
      p--;
}
x=x&~t;
t=t&y;
x=x|t;

first write the given number in it's binary format.
am considering only 8 bit at a time , 
5- 0000 0101
step 1: convert this binary format to one's compliment (ie)
change 0 to 1 and 1 to 0.
1111 1010

step 2: add 1 to the above binary format.

1111 1010
        1
-----------
1111 1011

so this binary format will get stored in memory and the sign
bit will be set to 1.

always the negative number will be stored only in 2's
compliment..........


thank u

Or in other words
~5+1

ans: +5

-5 will be stored as 2's complement of 5
2's complement of -5, is thus 2 times 2's complement of 5.
so the ans is 5 itself.

/*
 * Simple tree node representation
  */
struct node_t {
  struct node_t *left;
  struct note_t *right;
};

/*
 * Return the maximum depth of the tree given a pointer
 * to its root node.
 */
unsigned int
tree_depth (node_t *root)
{
   return (NULL == root) ? 0 :
           MAX(tree_depth(root->left, root->right)+1);
}

int n=strlen(x) // where x is pointer to hex string
int sum=0;
int leftshift=0;
while(n>0)
{
  if((x[n-1]>='0') && (x[n-1]<='9'))
    sum+=(x[n-1]-'0')<<leftshift;
  if((x[n-1]>='A') && (x[n-1]<='F'))
    sum+=(x[n-1]-'A'+10)<<leftshift;
  if((x[n-1]>='f') && (x[n-1]<='f'))
    sum+=(x[n-1]-'a'+10)<<leftshift;
  n--;
  leftshift+=4;
}

int n=strlen(x) // where x is pointer to hex string
int sum=0;
int leftshift=0;
while(n>0)
{
  if((x[n-1]>='0') && (x[n-1]<='9'))
    sum+=(x[n-1]-'0')<<leftshift;
  if((x[n-1]>='A') && (x[n-1]<='F'))
    sum+=(x[n-1]-'A'+10)<<leftshift;
  if((x[n-1]>='a') && (x[n-1]<='f'))
    sum+=(x[n-1]-'a'+10)<<leftshift;
  n--;
  leftshift+=4;
}

Bit fields can only be declared inside a structure or a
union, and allow you to specify some very small objects of a
given number of bits in length. Their usefulness is limited
and they aren't seen in many programs, but we'll deal with
them anyway.

Debuggers allow a user to halt a program and examine its
state. The debugger stops execution when some user-specified
condition is satisfied: Code breakpoints halt program
execution when a particular instruction is executed. Data
breakpoints halt program execution when a variable is
referenced. Code breakpoints are supported directly in
hardware on most machines and are fast. Data breakpoints,
however, are notoriously slow.

Even when the application is working fine, we need to 
ensure the performance of the application is good.
Hence performance testing has to be done.

Database testing is required to ensure the there is no constraint violence in the db. i.e while deleting any data, its not deleting any other data which is not supposed to be deleted. Again while adding we have to make sure that other than the expected tables, all tables remains intact.

I think the question says that , there are 2 diff combo 
boxes with the same name , Now how to parametrize them ?

as far as parametrization is concerned it can be done 
easily once the objects are diffrentiated .For the objects 
to be diff. recognized by QTP. You can add properties to 
the objects eg. index,size,hieght,width etc to difff. And 
then apply any parametrization u like.

Try the insert outpot value feature of QTP. That is 
designed for such scenarios only.

Please let me know if it helps.

Regards,
Jay Prakash

M

Madurai .......... starting character is M

madurai

madurai

madurai

Madurai

MADURAI. 'M' is the starting character


thank u

M
Madurai

the word is
MADURAI NO OTHER THAN

Madurai is the city name..
And starting letter is M..

MADURAI.'M' would be the starting character

answer is madurai

n&0x11111110

#define TURN_OFF_BIT_LAST(x) ((x) & ~1)

1)Give 5 reasons why a build can be released with bug
A:The build is not released with bugs.To know whether the
application is working according to requirements the build 
is tested.but intentionally the build is not released with 
bugs.
reasons may be:
1.poor requirements
2.developers may not have preior experience
3.mis communication between the team 
4.changing requirements
5.time pressure
6.software complexity

Build can be released with bug only with the following 
scenarios.
1 Bug severity is not effecting the functionality of 
product.
2 Test team has got the update that in the next build this 
bug will be fixed.
3 Time constraint.

1) 

a) Bug is of low severity and an alternate is available.
b) The module containing bug is isolated from the intended 
users.
c) Bug is not on priority for the BAs to be resolved in 
that release cycle.
d) Bug is cosmetic and of low priority.
e) Time constraing unless the bug is a major OR critical 
one.

2) 

Application: Word processor program i.e. WPP
Severty: Critical
Priority: Major
Platform: Microsoft XP (SP 3)
Reproducibility: Everytime

Issue:  WPP which crashes when the contents of 89th column 
is been sorted in assending order.

Steps to Reproduce:
1. Start  WPP
2. Add 100 columns
3. Add 2 rows
4. Add data in rows
5. Apply Ascending filter on row 89th by navigating Data 
tab > Filter
6. Click on Filter icon button on 89th column
7. Once data in the colmn is arranged in Descending order, 
again click on Filter icon button on 89th column to have 
data in ascending order

Instead of re arranding data, WPP crashes.

3)

1. Try to add a student for 15 for class 1-9, it should be 
added.
2. Try to add a student for 0 character for class 1-9, It 
should not be saved as name is required.
3. Try to add a student for 27 character for class 1-9 and 
validate truncation of name upto 26th character.
4. Try to add a student for 1 character for class 1-9, It 
should be added
5. Try to add a student for 26 character for class 1-9, It 
should be added
6. Try to add a student for 1-26 characterm for class 0, It 
should not be added
7. Try to add a student for 1-26 character for class 1, It 
should be added.
8. Try to add a student for 1-26 character for class 5, It 
should be added.
9. Try to add a student for 1-26 character for class 9, It 
should be added.
10. Try to add a student for 1-26 character for class 10, 
It should NOT be added as there is no class 10th.


Duplicacy:

While saving duplicate data for same class, system should 
alert user to save the name in a different format? But 
since it is character type only. how would we do that? May 
be DOB, Mother's maiden name

Student leave the program

11. Validate that student Name field Against the Class 1-9.

Removing duplicate data? System should alert by giving 
details like Father Name, DOB, Mother's maiden name to 
confirm deletion

12. Student gets promotion....

1. Class 1-8 gets promotion: Validate if class is 
incremented by 1.
2. Class 1-8 gets promotion: Student name is entered 
against incremented class.
3. Class 1-8 gets promotion: Name is deleted from previous 
class for the student who is just promoted.
4. Class 9 student gets promotion: Validate if student Name 
is deleted from class 9 who is just promoted.

4)

ddmmyy Boundry value Analysis Negative

0/0/1599 > Not Allwoed
0/0/10000 > Not Allwoed
0/0/1700 > Not Allwoed
0/1/1700> Not Allwoed
31/1/1700 > Not Allowed
1/0/1700 > Not Allwed
30/13/1700 > Not Allowed
31/13/1700 > Not Allwoed
0/12/1700 > Not Allowed
30/12/1700 > Allowed and validate if it prints in 31/12/1700
30/10/1599 > Not Allowed
30/10/10000 > Not Allowed

0/0/1599 > Not Allwoed
0/0/10000 > Not Allwoed
0/0/1700 > Not Allwoed
0/1/1700> Not Allwoed
31/1/1700 > Not Allowed
1/0/1700 > Not Allwed
30/13/1700 > Not Allowed
31/13/1700 > Not Allwoed
0/12/1700 > Not Allowed
30/12/1700 > Allowed and validate if it prints in 31/12/1700
30/10/1599 > Not Allowed
30/10/10000 > Not Allowed

Hi,

I am aware of these questions thanks for posting and these 
are very helpful , one query can anybody post latest 
question asked in August 2009 in Adobe 

Thankful to all!!!!!

satijaa.tarun@gmail.com

Regards

Tarun