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Different types of Pointers

| Wednesday, 18 December 2013
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Generic pointer: void pointer in c is known as generic pointer. Literal meaning of generic pointer is a pointer which can point any type of data. Example: void *p;
   Wild pointer------------------- A pointer in c which has not been initialized is known as wild pointer.

Near pointer ------------------- The pointer which can points only 64KB data segment is known as near pointer. Size of near pointer is two byte. Near pointer only hold 16 bit offset address. Near pointer cannot access beyond the data segment like graphics video memory, text video memory etc. With help of keyword near, we can make any pointer as near pointer. eg. int near * p;   

Dangling pointer----------------- If any pointer is pointing the memory address of any variable but after some variable has deleted from that memory location while pointer is still pointing such memory location. Such pointer is known as dangling pointer. 

Far Pointer --------------- ------------ The pointer which can point or access whole the residence memory of RAM (Generally 1 GB = 64KB * 16 segment ) as far pointer. Size of far pointer is 4 byte or 32 bit. First 16 bit stores: Segment number Next 16 bit stores: Offset address %Fp is used for print offset and segment address of pointer in printf function in hexadecimal number format. eg. int far * p;

Huge Pointer Warning !!!! ----------- If you will increment huge pointer it will increment both offset and segment address unlike to far pointer which only increments offset address. So if you have little knowledge about huge pointer and you are using huge pointer then you can easily access and modify the IVT, device driver memory, video memory etc. This might be dangerous for your computer.

Different types of Pointers

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GATE QUESTIONS

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All digital circuits can be realized using only
(a) Ex-OR gates (b) Multiplexers
(c) Half adders (d) OR gates

Answer B
Because multiplexers and decoders are called "universal Logic " so using these we can make any circuits

GATE QUESTIONS

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Founders of UNIX

| Tuesday, 17 December 2013
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Ken Thompson and Dennis Ritchie

Founders of UNIX

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MOST FAMOUS ANDROID 'S FOUNDERS

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Founders of Android
There are total four founders of android
named Andy Rubin, Rich Miner, Nick Sears and Chris White

MOST FAMOUS ANDROID 'S FOUNDERS

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FOUNDERS OF IMPORTANT COMPANIES

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Founder of Apple Computers – Steve Jobs
Founder of Artificial Intelligence – John
McCarthy
Founder of Bluetooth – Ericsson
Father of Computer – Charles Babbage
Father of ‘C’ Language – Dennis Ritchie
Founder of Email – Shiva Ayyadurai
Founder of Google – Larry Page and Sergey
Brin
Founder of Internet – Vint Cerf
Father of ‘Java’- James Gosling
Father of JQuery – John Resig
Founder of Keyboard – Christoper Latham
Sholes
Founder of Microsoft – Bill Gates and Paul
Allen
Founder of Mobile Phones – Martin Cooper
Founder of Mouse – Douglas Engelbart
Founders of Oracle – Ed Oates, Larry Ellison,
Bob Miner
Founder of Php – Rasmus Lerdorf
Founder of USB – Ajay V.Bhatt
Founder of WWW – Tim Berners-Lee
Founder of Yahoo – Jurry Yang and David Filo..

FOUNDERS OF IMPORTANT COMPANIES

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HOW TO FIND CYCLOMATIC COMPLEXITY

| Monday, 16 December 2013
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Consider the following pseudo-code :
If (A > B ) and (C > D) then
A = A + 1
B = B + 1
Endif
The cyclomatic complexity of the pseudo-code
is
(A) 2
(B) 3
(C) 4
(D) 5
ANSWER :3

V(G)=p+1 where p is the no of predicates.Here
we have two predicates or two conditions.ther
efore 2 + 1 =3

HOW TO FIND CYCLOMATIC COMPLEXITY

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GATE QUESTIONS AND SOLUTIONS

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Consider a logical address space of 8 pages of
1024 words mapped with memory of 32
frames.
How many bits are there in the physical
address ?
(A) 9 bits
(B) 11 bits
(C) 13 bits
(D) 15 bits

size of logical address space is No. of pages *Page size = 8 * 1024 = 2^3 * 2 ^10 = 2^13
No. of bits for logical address is 13
Size of Physical address space is 2^5(32) * 2^10 (1024)=
2^15 No. of bits for physical address is 15

GATE QUESTIONS AND SOLUTIONS

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PRE ORDER POST ORDER IN ORDER

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The post order traversal of a binary tree is
DEBFCA. Find out the pre- order traversal.
(A) ABFCDE
(B) ADBFEC
(C) ABDECF
(D) None of the above
Ans:-C
Explanation:-
A binary tree is a data structure which has at
most two child nodes. A binary tree can be
traversed in three ways. Inorder, preorder and
postorder. The three methods of traversal can
be explained this way.
TYPE FIRST SECOND THIRD
Inorder L - Left Ro - Root R - Right
Preorder Ro - Root L - Left R - Right
Postorder L - Left R - Right Ro - Root
Left will always be followed by right. But Root
would be either in the front or in the middle
or at the last depending on whether it is pre,
in or postorder traversal. So given a post order
traversal of a binary tree, we can know the
root first. In the question, the post order
traversal is given as DEBFCA. Since Root is the
last node to be traversed in a post order
traversal we know one thing for sure. A is the
root.
Next, we are left with only DEBFC. Here some
of the nodes belong to the left side of the
binary tree and some belong to the right side.
How many nodes belong to the left and how
many belong to the right. Since left side of the
binary tree is considered first, and since every
node is expected to have at most two child,
DEB will be the left side of the binary tree and
FC would be the right.
Now, we know that FC is in the right side of
the binary tree. Again the last node would be
the root of the sub tree and F its left side.
Next we come to the left side of the binary
tree and it is DEB. Again B would be the root
of the sub tree. D and E are its left and right
side respectively. So the binary tree would
look something like this given below.
After constructing the binary tree, writing the
preorder traversal is very simple. In preorder
traversal root comes first. Since A is the root,
A would appear first. Following Root would be
the Left and Right sub tree and so the left
subtree would be BDE. Again B would be the
root of the left sub tree followed by D and E
which are the left and right child respectively.
SO the preorder traversal till now would be
ABDE. Last comes the Right sub tree. Here
again C would be the root of the right sub tree
followed by C its left child. So the entire
preorder traversal of the tree would be
ABDECF. So option C is the right answer.

PRE ORDER POST ORDER IN ORDER

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GATE QUESTIONS AND SOLUTIONS

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Given memory partitions of 100 K, 500 K, 200
K, 300 K and 600 K (in order) and processes of
212 K, 417 K, 112 K, and 426 K (in order),
using the first-fit algorithm, in which partition
would the process requiring 426 K be placed ?
(A) 500 K
(B) 200 K
(C) 300 K
(D) 600 K
Answer d
Because we choose linear method of first fit 1st 212 k will fit into 500k space and remaining space is 288k.and then 417k fit into 600k remaining space is 183k and 112k process is fitted into remaining 183k memory. Then 426k process doesnt have space so linearly choose space so it uses 500k partition and 212k process fitting into remaining space of 288k a new partition

GATE QUESTIONS AND SOLUTIONS

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Incredible watch cell phone

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TW206 Watch Cell
Phone with Quad Band Single SIM 1.5 inch
HVGA Touch Screen Camera Bluetooth (Black)

Highlights:
Type: Watch cell phone
Color: Black
Battery: 400 mAh
Network frequency: GSM 850/900/1800/
1900MHz
Single SIM, Single standby
Single camera
Bluetooth supported
MP3/MP4 function supported

Incredible watch cell phone

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GATE QUESTION CS 2005

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Consider the following C-program:
void foo( int n, int sum)
{
int k = 0, j = 0;
if (n == 0) return;
k = n % 10;
j = n / 10;
sum = sum + k;
foo (j, sum);
printf ("%d," , k);
}
int main ()
{
int a = 2048, sum = 0;
foo (a, sum);
printf ("%d\n" , sum);
getchar ();
}
What does the above program print? (GATE
CS 2005)
(a) 8, 4, 0, 2, 14
(b) 8, 4, 0, 2, 0
(C) 2, 0, 4, 8, 14
(d) 2, 0, 4, 8, 0
Answer (d)
sum has no use in foo(), it is there just to
confuse. Function foo() just prints all digits of
a number. In main, there is one more printf
statement after foo(), so one more 0 is printed
after all digits of n.

GATE QUESTION CS 2005

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TRANSPARENT 3D COMPUTER

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A transparent computer that allows users to
reach inside and touch digital content has been
unveiled at the Technology, Entertainment and
Design (TED) conference in Los Angeles.
TED fellow Jinha Lee has been working on the
SpaceTop 3D desktop in collaboration with
Microsoft.
Allowing people to interact with machines in the
same way they do with solid objects could make
computing much more intuitive, he told the
BBC.
He can see the system coming into general use
within a decade.
The system consists of a transparent LED display
with built-in cameras, which track the user's
gestures and eye movements.
The design was inspired by what he sees as a
human need to interact with things.
"Spatial memory, where the body intuitively
remembers where things are, is a very human
skill," he said.
Translating this to the digital world will enable
people to use computers more easily as well as
complete more complex tasks.
"If you are working on a document you can pick
it up and flip through it like a book," he said.
For more precise tasks, where hand gestures are
not accurate, there is a touchpad. It will allow,
for example architects to manipulate 3D models.

TRANSPARENT 3D COMPUTER

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Asus Transformer Book Trio: a hybrid laptop

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Asus Transformer Book Trio: a hybrid laptop
that runs Windows and Android
Asus has launched the Transformer Book Trio in
India, a unique 11.6-inch 'three-in-one'
ultraportable device. The device costs Rs 98,099
(USD 1615) and is available across select Asus
authorised retail outlets later in December 2013.

Asus Transformer Book Trio: a hybrid laptop

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