Computer Science

yes

http://www.osix.net/modules/article/?id=275

muzzy writes "Here's some code for you kids. It demonstrates the concept of stack, implemented with a linked list mechanism to support virtually infinitely large stack sizes. Happy reading."

/* Simple Dynamically Allocating Stack Implementation in C
*
* Copyright (C) 2002 Muzzy of Worst Coders
*/

A class is the definition of a type -- it consumes no memory in an off itself. An object is an instance of a class, and therefore consumes memory, equal to the total size of all its member variables (attributes), including base class members, plus padding for alignment. If the class declares any virtual methods, a v-table is also created, which consumes additional memory.

FAT32 is a widely recognized file system. While not necessarily the most modern or most reliable system, using FAT32 per default allows manufacturers to sell one version for use with many computers and operating systems.

The same cannot be said from more advanced file systems such as NTFS or Reiser, to name just two.

Expert users can always re-format a device with the file system of choice prior to first use.

The term 'domain' is too general to compare to the idea of a forest. A domain and the AD can be a part of a forest. This includes; domain controllers, child domains, domain functionality, replicators, directory service and so on. The concept of creating a forest was first introduced in the windows 2003 AD architecture. Suffice to say interoperability with server 2000 and NT (which do not recognize the forest) poses limitations and security issues. Hence four levels of functionality. Some are, in my opinion, basically unsound with regards to the security levels of a forest. A forest is not to be taken lightly. It requires much research and preparation. The term 'domain' applies across the board in a forest. Moreover, a forest relies on security. The PC you start the first installation of a forest will be considered the root and will hold the high level admins such as the enterprise and schema admins. Making forest trusts (only on root domain) facilitates communications between domains and ADs that share the same SPN (service principle name) which have to be resolved at a remote location in another forest. The configuration also requires IAS, Kerberos, UPN, SPD, SID namespaces .... What am I forgetting? Thinking about configuring the root forest on the first PC makes you dizzy with abbreviations acronyms, protocols, group security, etc ... Comprehensive research and planning are crucial. Managing forests and domain is hard enough as it is. I'd say this basic principle of security properties could be considered the largest difference between a 'forest' and a 'domain'.

Calculate the amount of additional memory used by the algorithm relative to the number of its inputs. Typically the number of inputs is defined by a container object or data sequence of some type, such as an array. If the amount of memory consumed remains the same regardless of the number of inputs, then the space complexity is constant, denoted O(1) in Big-Omega notation (Big-O). If the amount of memory consumed increases linearly as n increases, then the space complexity is O(n).

For example, the algorithm that sums a data sequence has O(1) space complexity because the number of inputs does not affect the amount of additional memory consumed by the accumulator. However, the algorithm which copies a data sequence of n elements has a space complexity of O(n) because the algorithm must allocate n elements to store the copy.

Other commonly used complexities include O(n*n) to denote quadratic complexity and O(log n) to denote (binary) logarithmic complexity. Combinations of the two are also permitted, such as O(n log n).

What are the defects in a crystal? Describe them with a neat diagram

A process or thread holding a resource while waiting to get hold of another resource

Thousands if not tens or hundreds of thousands.

Let p and q be the two polynomials represented by the linked list. 1. while p and q are not null, repeat step 2. 2. If powers of the two terms ate equal then if the terms do not cancel then insert the sum of the terms into the sum Polynomial Advance p Advance q Else if the power of the first polynomial> power of second Then insert the term from first polynomial into sum polynomial Advance p Else insert the term from second polynomial into sum polynomial Advance q 3. copy the remaining terms from the non empty polynomial into the sum polynomial.

George H. Mealy presented the concept of a finite-state machine that became known as the Mealy machine in a 1955 paper, "A Method for Synthesizing Sequential Circuits." This machine differs from a Moore machine in that rather than the outputs being defined by the current state, the outputs are defined by both the current state and the current inputs.

no

The sum of 20 and -4 is +16, while the product of 4 and -2 is -8. The answer to the question is therefore -2.

// Indexed Allocation

#include

#include

#include

struct node {

int file_name;

int data;

int is_free;

int size;

int directory;

int cnt;

struct node* link;

struct node* inner_link[30];

};

struct node* defaultFile() {

struct node* temp = (struct node*)malloc( sizeof(struct node) );

temp->link = NULL;

temp->is_free=0;

temp->data='a';

temp->file_name=99;

temp->directory = 1;

return temp;

}

struct node* insert(struct node *rt, int size, int fname) {

struct node* temp = rt;

struct node* first=NULL,*last=NULL,*tt=NULL;

int flag=0,act_size=0, inner_fname=100,cnt=0;

last=first;

act_size = size;

if ( size>temp->size ) {

printf("There is not enough space on the disk to write that file\n");

return rt;

}

while( temp->link ) {

temp = temp->link;

}

first = defaultFile();

while( act_size>0 ) {

tt = defaultFile();

if (act_size>50)

tt->size = 50;

else

tt->size = act_size;

tt->file_name = inner_fname;

first->inner_link[cnt] = tt;

tt->directory = 0;

tt->is_free = 0;

act_size -= 50;

inner_fname += 1;

cnt += 1;

}

temp->link = first;

first->is_free = 1;

first->cnt = cnt;

first->size = size;

first->file_name = fname;

act_size = rt->size;

rt->size = act_size-size;

return rt;

}

void printFiles(struct node* rt) {

struct node *temp = rt, *tt;

int first=0,cnt=0;

printf("format is (File name, size)\n");

printf("\t(%d,%d)\n",rt->file_name,rt->size);

while( temp ) {

if ( temp->is_free ) {

printf("\t(%d,%d)\n",temp->file_name,temp->size);

first = 0;

while( cntcnt ) {

tt = temp->inner_link[cnt];

printf("\t\t(%d,%d)\n",tt->file_name, tt->size);

cnt += 1;

}

printf("\n");

}

temp = temp->link;

}

}

struct node* combine(struct node *rt,int fname) {

struct node *temp=rt,*nt=NULL,*temp1=temp->link;

int size=0;

if ( rt->file_name==fname ){

printf("You cannot that file as thats just to show that that much amount of space is left in the disk\n");

return rt;

}

while( temp1 ) {

if (temp1->is_free==0 && temp1->file_name==fname ) {

size = temp1->size;

temp->link = temp1->link;

temp1 = temp->link;

}

else {

temp = temp1;

temp1 = temp1->link;

}

}

rt->size += size;

return rt;

}

struct node* deleteFiles(struct node* rt, int fname) {

struct node *temp = rt,*nt=NULL;

int flag=0;

while( temp && flag==0 ) {

if (temp->file_name==fname) {

temp->is_free=0;

flag=1;

}

temp = temp->link;

}

if( flag==0 ){

printf("There doesnt exist any file with that name\n");

}

return combine(rt,fname);

}

int main() {

int flag,no,size,data;

struct node *root;

root = defaultFile();

root->size=1000;

data=100;

flag=no=size=0;

while( flag==0 ) {

printf("Enter no's \n1.insert\n 2.Delete\n 3.Print files \n 4.Exit\n");

scanf("%d",&no);

printf(" no is %d\n",no);

switch(no) {

case 1:

printf("Enter file size\n");

scanf("%d",&size);

root = insert(root, size, data);

data = data+1;

break;

case 2:

printf("Enter file name to delete\n");

scanf("%d",&size);

root = deleteFiles(root, size);

break;

case 3:

printFiles(root);

break;

case 4:

flag=1;

printf("Quitting from loop\n");

break;

default:

printf("Enter a valud no \n");

break;

}

}

}

In a passage of text, individual words and punctuaton marks are called tokens.the smallest individual unit in a prog is called a token..

There are many tools you can use to create design, Type Tool to create text, Shape Tools to draw shapes, Pen Tool to draw shapes, Brush Tool to paint and they are located on left side of Screen in Toolbox.

You have to pass the ADM201 exam.

This examination cost $200 and you can register through webassessor.com

Passing it though is not that easy, although all questions are covered in the help and training of Salesforce which is available free in their website, most aspirants still have to enroll a $3000 course on Administration Essentials - yet, not all of them always pass.

You have to review a lot on your own. See sources below.

1. Find the element is the list. Let pointer 'p' point to it.

2. Delete it from the list:

if (p->Prev) p->Prev->Next = p->Next;

else List->First = p->Next;

if (p->Next) p->Next->Prev = p->Prev;

else List->Last = p->Prev;

3. Release the memory associated with it.

You get a screen recorder and 3 other high levels and record you simming once and it does it over and over