C Programming

Searching is slower for linked-lists rather than (ordered) trees because you need to iterate through each element serially, whereas for an ordered (binary) tree you use a "divide and conquer", otherwise known as binary search, method.

Think of it this way. Think of a number between 1 and 128. Perhaps the number is 97. With an ordered list, it would take 97 comparisons to find the item. With an unordered list, it would take an average of 64 comparisons - if the order was random. With a binary tree, i.e. a binary search, you only need 7 comparisons. In fact, you never need more than 7 comparisions for a tree size of 128, and on average, you would use less than that.

There are a few ways of going about this. The easiest way would be through a literal string:

printf("X X X X X X X X X");

If you want to print an arbitrary number of X's (in this case, 9), a for() loop would work:

for (count=0; count<9; count++) printf("X ");

This code would generate a final space at the end. To fix this and only use spaces to separate X's:

for (count=0; count<9; count++) printf("%sX", count?" ":"");

#include<stdio.h>

#include<conio.h>

void main()

{

int n,a=2;

clrscr();

printf("\n enter n");

scanf("%d",&n);

if(i<=a-1)

{

if(a%i==0)

{

printf("\n the given number is not a prime number");

break;

}

i++;

if(a==i)

{

printf("\n the given number is a prime number");

}

getch();

output:

enter the value of n:2

the given number is prime number

it will give the substring of length(as per the user) from the actual string,and the starting length from where it has to copy the substring from actual should be given by the user.

binary language

Design of the database (Database Design) refers to a given application environment, optimize the structure of the database, the database and applications, which can efficiently store data to meet the application needs of various user information needs and processing requirements). At www. myelibrary.net.In you can clearly understand what is the database design.

In a segment of memory, whose name is 'data segment'

A type system is generally very useful. It prevents you from making many mistakes. For example adding 1 to "cool" would result in a garbage answer but you might do it accidentally without types.

However types are sometimes constraining and inflexible. Perhaps I really do want to add 1 to 'c' because I know exactly what new ascii value it would give me. Type systems also increase compile time, run time, or both.

Note: You may be confusing no typing with dynamic typing. With dynamic typing you generally do not label a variable with a type in its declaration. Scripting languages like python are often like this. However they remain strongly typed languages. Assembly is an example of a untyped or one type language.

No. Generally, one instruction in a high level language corresponds to many instructions in machine language.

in dynamic list after adding a new element size of the list disturbed so deletion and insertation is necessary to add a new element

Yes, please do. Here is an example:

#include <stdio.h>

int main (void)

{ puts ("Hello, World"); return 0; }

A library file on a computer is the user's personal file. It could be a series of pictures downloaded to the computer, documents scanned into the computer, or a music file.

C is ideally suited to low-level programming, but has been used to write operating system kernels, device drivers, subsystems programming and applications software of all types. C++ now dominates across all these domains as it is capable of producing far more efficient machine code much more easily. However, even C++ programmers will make use of C-style programming in combination with its object-oriented and generic programming paradigms.

Possible.

void foo (void)

{

int array [10][20];

...

}

Temporary files are used a lot in computers. Most of the time, a temporary file is created when changes to a permanent file cannot be made until a certain process is complete.

A simple analogy might be painting a wall. Before you can paint the wall, anything hanging on the wall must be moved to a temporary holding area while the wall is being painted. After the painting process is complete, you move the items from their temporary location back to the permanent location - the wall that they hang on.

Computers actually follow a very similar process whenever a file is updated. Often times, the original file is copied to a temporary folder where the computer will make any changes needed to the file. When all the changes are complete, the computer might request your approval before the information is permanently stored in the permanent location where the file belongs.

Many modern word processing software programs (like Microsoft Word, Corel Wordperfect, or OpenOffice Writer) will create temporary files for new documents before the user actually saves the document to its permanent folder. This way, if the power fails suddenly, many of these programs can then check for any temporary files (such as a 30 page essay you've been working on for the past three hours) and "restore" your work. When you actually tell the program to save the file and give it a file name, the file is then moved to its permanent location.

There are a number of requirements to be a heavy machine operator. Depending on the type of machinery that is being operated, firstly a good level of health is required as the place of work could require some minimal lifting and carrying of smaller goods, in addition to this a college or apprenticship scheme may need to have been completed as will a commercial drivers license test that sometimes has a fee attatched to it.

/* program to study Reader/Writer problem with Semaphore along with output*/

#include<stdio.h>

#include<pthread.h>

#include<semaphore.h>

void reader1();

void writer1();

void reader2();

void writer2();

sem_t rsem;

sem_t wsem;

sem_t x,y,z;

int rc=0, wc=0,p=0;

int main()

{

int r,w,i=0,ch=2;

pthread_t t1[5],t2[5];

printf ("\n\nEnter the number of READERS: ");

scanf ("%d", &r);

printf ("\n\nEnter the number of WRITERS: ");

scanf ("%d", &w);

printf ("\n\nMENU\n\n\t1.READERS HAVE PRIORITY");

printf ("\n\n\t2.WRITER HAVE PRIORITY");

printf ("\n\n\tENTER YOUR CHOICE:");

scanf ("%d", &ch);

sem_init(&rsem, 0, 1);

sem_init(&wsem, 0, 1);

sem_init(&x, 0, 1);

sem_init(&y, 0, 1);

sem_init(&z, 0, 1);

if (ch==1)

{

for (i=0;i<r;i++)

pthread_create(&t1[i], NULL, (void *)reader1, (void *)i );

for (i=0; i<w; i++)

pthread_create(&t2[i], NULL, (void *)writer1, (void *)i);

//pthread_create(&t1[w-1], NULL, (void *)reader1, (void *)(w-1) );

//pthread_create(&t2[r-1], NULL, (void *)writer1, (void *)(r-1) );

for (i=0;i<r;i++)

pthread_join(t1[i], NULL);

for (i=0;i<w;i++)

pthread_join(t2[i], NULL);

}

else if (ch==2)

{

//pthread_create(&t2[0], NULL, (void *)writer2, 0);

//pthread_create(&t1[0], NULL, (void *)reader2, 0);

for (i=0;i<r;i++)

pthread_create(&t2[i], NULL, (void *)reader2, (void *)i );

for (i=1;i<w;i++)

pthread_create(&t1[i], NULL, (void *)writer2, (void *)i );

//pthread_create (&t2[r-1], NULL, (void *)writer2, (void *)(r-1) );

//pthread_create (&t1[w-1], NULL, (void *)reader2, (void *)(w-1) );

for(i=0; i<r; i++)

pthread_join(t1[i], NULL);

for(i=0; i<w; i++)

pthread_join(t2[i], NULL);

}

else

printf ("\nINVALID CHOICE....PROGRAM TERMINATED.");

sem_destroy(&rsem);

sem_destroy(&wsem);

return 0;

} // main function close

void reader1 (void *i)

{ int i1;

for(i1=0;i1<10;i1++)

{

sem_wait(&x);

rc++;

if (rc==1)

sem_wait(&wsem);

sem_post(&x);

printf ("\n\t \t \t \tREADER %d IS READING NOW %d", (int)i,p);

sem_wait(&x);

rc--;

if(rc==0)

sem_post(&wsem);

sem_post(&x);

sleep(2);

//printf ("\nREADER %d HAS FINISHED READING\n");

}

}

void writer1(void *i)

{ int i1;

for(i1=0;i1<10;i1++)

{ sem_wait(&wsem);

p++;

printf ("\nWRITER %d IS WRITING NOW %d", (int) i,p);

sem_post(&wsem);

if ((int)i % 2 == 0)sleep(1); else sleep(6);

// printf ("\nWRITER %d HAS FINISHED WRITING\n");

}

}

void reader2(void *i)

{

sem_wait(&z);

sem_wait(&rsem);

sem_wait(&x);

rc++;

if(rc==1)

sem_wait(&wsem);

printf ("\nREADER %d IS READING NOW", (int)i);

sem_post(&x);

sem_post(&rsem);

sem_post(&z);

sleep(1);

sem_wait(&x);

rc--;

if (rc==0)

sem_post(&wsem);

sem_post(&x);

//printf ("\nREADER %d HAS FINISHED READING\n");

}

void writer2(void *i)

{

sem_wait(&y);

wc++;

if (wc==1)

sem_wait(&rsem);

sem_post(&y);

sem_wait(&wsem);

printf ("\nWRITER %d IS WRITING NOW", (int) i);

sem_post(&wsem);

sem_wait(&y);

wc--;

if(wc==0)

sem_post(&rsem);

sem_post(&y);

sleep(1);

//printf ("\nWRITER %d HAS FINISHED WRITING\n");

}

/*

student@student-desktop:~/Documents$ ./rdwrt

Enter the number of READERS: 4

Enter the number of WRITERS: 4

MENU

1.READERS HAVE PRIORITY

2.WRITER HAVE PRIORITY

ENTER YOUR CHOICE:1

READER 0 IS READING NOW

READER 1 IS READING NOW

READER 2 IS READING NOW

READER 1 IS READING NOW

READER 2 IS READING NOW

READER 3 IS READING NOW

READER 1 HAS FINISHED READING

READER 0 HAS FINISHED READING

READER 1 HAS FINISHED READING

READER 2 HAS FINISHED READING

READER 2 HAS FINISHED READING

READER 3 HAS FINISHED READING

WRITER 0 IS WRITING NOW

WRITER 0 HAS FINISHED WRITING

WRITER 1 IS WRITING NOW

WRITER 1 HAS FINISHED WRITING

WRITER 2 IS WRITING NOW

WRITER 2 HAS FINISHED WRITING

WRITER 3 IS WRITING NOW

WRITER 3 HAS FINISHED WRITING

student@student-desktop:~/Documents$ ./rdwrt

Enter the number of READERS: 4

Enter the number of WRITERS: 4

MENU

1.READERS HAVE PRIORITY

2.WRITER HAVE PRIORITY

ENTER YOUR CHOICE:2

WRITER 0 IS WRITING NOW

WRITER 0 HAS FINISHED WRITING

WRITER 1 IS WRITING NOW

WRITER 1 HAS FINISHED WRITING

WRITER 2 IS WRITING NOW

WRITER 2 HAS FINISHED WRITING

WRITER 3 IS WRITING NOW

WRITER 3 HAS FINISHED WRITING

READER 1 IS READING NOW

READER 0 IS READING NOW

READER 2 IS READING NOW

READER 3 IS READING NOW

READER 2 HAS FINISHED READING

READER 1 HAS FINISHED READING

READER 0 HAS FINISHED READING

READER 3 HAS FINISHED READING

*/

Written by: Fabianski Benjamin

There is insufficient information in the question to properly answer it. You did not provide the example "here". Please restate the question.

Object oriented programming and structured programming.

#include<stdio.h> #include<conio.h> void main() { clrscr(); int x=0; do { printf("Kaushal"); printf("\n"); x++; }while(x<10); getch(); }

Borrowing the isPrime function from another answer of mine. Note that this will result in terrible performance for large arrays of numbers. (You should look into one of the sieve algorithms for this)

// returns 1 if n is prime, 0 otherwise

void isPrime(const int n) {

// We know that if n is composite, then we will find a factor in the range [2,sqrt(n)]

// so we compute the square root only once to limit our number of calculations.

const int sqrt_n = sqrt(n);

// Iterate through possible factors

int i;

for( i = 2; i <= sqrt_n; ++i ) {

// If n is divisible by i (n%i==0) then we have a factor

if(!(n % i)) {

return 0;

}

}

// If we get here, we know n has no factors other than itself and 1

return 1;

}

// returns the sum of all prime numbers in nums

int findPrimeSum(const int numsLength, const int[] nums) {

int sum = 0;

// iterate through nums and add up all the primes

int i;

for(i = 0; i < numsLength; ++i) {

if( isPrime(nums[i] ) {

sum += nums[i];

}

}

return sum;

}

Its Unicode value is 221A according to System tool Character map Advanced view Unicode subrange Math operators.

But I haven't done C in awhile, so I don't know how to or if you can.

ASCII value of root symbol is 251. In C we can print this symbol by printing the character value as below

printf("%c",251);

this will print the root symbol