#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int happyNumber(int n)
{
    char temp[10];
    itoa(n,temp,10);
    int len=strlen(temp);
    int ret,sum=0,i;
    
    if(n==1)
         return 1;
    else if(n==4)
         return 0;
    else
    {
            for(i=0;i<len;i++)
                         sum += (temp[i]-48)*(temp[i]-48);
            ret = happyNumber(sum);
    }
    
    return ret;
}


int main()
{
  int n;
  printf("Enter the number : ");
  scanf("%d",&n);
  if(happyNumber(n) == 1)
      printf("\n\n%d is a HAPPY NUMBER",n);
  else
      printf("\n\n%d is NOT A HAPPY NUMBER",n);
      
 fflush(stdin);
 getchar();
 return 0;
}

29

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
  int nocw,i,j,flag=0;
  char
cenWords[20][12],text[205],temp[12],finalText[450]={'\0'};

  printf("Enter number of words to be censor : ");
  scanf("%d",&nocw);

  printf("\nEnter words to be censor : ");
  for(i=0;i<nocw;i++)
                     scanf("%s",cenWords[i]);
                     
  fflush(stdin);
  printf("\nEnter text to censor : ");
  gets(text);
  
  for(i=0;i<strlen(text);i++)
  {
     j=0;
     flag=0;
     while(!(text[i]==' '||text[i]=='\t'||text[i]=='\n'))
     {
           temp[j++]=text[i++];
     }
     temp[j]='\0';

       for(j=0;j<nocw;j++)
       {
          if(strcmp(temp,cenWords[j])==0)
          {
            strcat(finalText,"<censored> ");
            flag=1;
            break;
          }
       }
       
       if(flag==0)
       {
                  strcat(finalText,temp);
                  strcat(finalText," ");
       }
  }    
  
  printf("\n\n :: FINAL TEXT :: \n\n");
  puts(finalText);
 getchar();
 return 0;
}

here it will print :  either NOT EQUAL  or EQUAL
according to me and my compailer it's NOT EQUAL only.......
depends upon the compailer actually......

here variable 'a' is assigned wit float and 'b' is assigned
with double.... 
here comes the problem ie. of allocation of no. of bytes

according to my compailer float allocates 4 bytes and double
allocates 8 bytes.... so when it compares each location bit
by bit the float will lag by two bytes of space than double
has .. so since the comparsion takes a decision tat it's not
equal since it dosent check the other two locations of
double wit float which is lagging wit that two bytes of
memory space....... so it prints NOT EQUAL


thank you

answer is not equal;
bcoz float cannot be compared with double.

It will print "not equal". Reason is the difference in 
precision of the numbers. ie numbers like 8.8 or 8.1 can't 
be stored with complete precision in binary sysetm since 
it's mantissa part will not end but continues with a 
series. So value calculated for single precision(float) 
number will be slightly different from the value calculated 
for double precision (double) number. To verify this use 
gcc and gdb in linux.

If you try with numbers 8.25, 8.5 or 8.75 the program will 
print "equal" since the mantissa part ends with in the 
precision.

I think this has nothing to do with compiler version or 
inabilty of comparision.

semaphores helps is a varible which is used to protect the shared data .hence avoiding the coruption of data by the threads acting simultaniously..

#include<stdio.h>
int main()
{
 char c = 'X';
 int i,n,sum = 0;
 printf("Enter values for n\n");
 scanf("%d\n",&n);
 for(i=0;i<n;i++)
 {
  sum = sum+power(c,i)/fact(i);
 }
 fact(int i)
 {
   int p=1 ,fact; 
   if(p<=i)
    {
     fact = p*fact(p++);
    } 
   return fact;
 }

bcoz i didn't get any other software company.

i want iearn everthing about softwares.

i want to be a part of it.As it is my pleasure.

tell related to R&D of the company, new learning , the 
culture,the policy.

bcoz i wanna learn somethig i course of my service with 
self satisfaction and this company is tthe best to give me 
that opertunity

To pursue a dynamic and challenging career with your 
esteemed organization of , it will give me value addition 
to my career  as well as will also offer opportunity to 
enhance my professional skills.

it's the one and one leading company.team quatination is very super and excellent.the team leader is help to the staff.
encorage the team members.

Null pointers returns null value and void pointer returns no
value

NULL POINTER :
the pointer which dosent point to any memory location is 
called NULL POINTER.

VOID POINTER :
the pointer is a one which can only point to a particular 
memory location of it's own type.... but when the pointer 
is given void it can be mde to point any location of 
different type

what is except handling in c sharp?

#include<stdio.h>
#include<conio.h>
void main()
{
int n;
printf("enter the lines :");
scanf("%d",&n);
for(int i=1;i<=n/2;i++)
{
printf("*\n");
for(int j=1;j<=2*i;j++)
printf("%d",j);
printf("\n");
}
if(n%2!=0)
printf("\n*");
getch();
}

thank u

#include<stdio.h>
#include<conio.h>
void main()
{
int n;
printf("enter the lines :");
scanf("%d",&n);
for(int i=1;i<=n/2;i++)
{
printf("*\n");
for(int j=1;j<=2*i;j++)
printf("*");
printf("\n");
}
if(n%2!=0)
printf("\n*");
getch();
}

void main() {
   int n = 5 ; // take use input here
   int i, j;
   for( i =1; i<=n; i++){
     printf("\n *");
     for(j=0; j<=i; j++){
        printf(" *");
     }
   }
}

#include<stdio.h>
int main()
{
int n,i,j;
printf("enter the lines :");
scanf("%d",&n);
for(i=1;i<=n/2;i++)
{
printf("*\n");
for(j=1;j<=2*i;j++)
printf("*");
printf("\n");
}
if(n%2!=0)
printf("\n*");

}

compiled with GCC

#include<stdio.h>
#include<conio.h>
main()
{
  int i,j;
  clrscr();
  for(i=1;i<=10;i++)
  {
     if(i%2!=0)
     printf("*\n");
     else
     {
      for(j=1;j<=i;j++)
      printf("*");
      printf("\n");
      }
    }
    getch();
}

void main() {
   int n = 5 ; // take use input here
   int i, j;
   for( i =1; i<=n; i++){
     printf("\n *");
     for(j=1; j<i; j++){
        printf(" *");
     }
   }
}

_start will call the main function in c under linux

When the programme is commpiled,in the kernel execl will be
called (execl will replace current process image with new
process image)in the new process strat up routine
ia executed this start up routine is written in C abd assembly.
In this routine exit(main()); function will called,this will
call to main() function.

Kiransing Paradeshi
Software Engineer
SPA Comuters (P) Ltd
Banglore,India
kiransing4u@gmail.com

#include <stdio.h>

void div(int n){
int i=2;
    while(n%i!=0 && i!=n){
         i++; 
    }
     printf("%d ",i);
	if(i!=n){
	  div(n/i);
	}

}

main(){
int i;
printf("Enter number:");scanf("%d",&i);
printf("1 ");
div(i);
}

void div(int n){
static int i=1;
    while(i<=n){
    if(!(n%i))printf(" %d",i);
         i++;
    }
}

main(){
int i;
clrscr();
printf("Enter number:");scanf("%d",&i);
div(i);
getch();
}

#include<stdio.h>
int main()
{
 int n,i=1;
 printf("Value for n\n");
 scanf("%d\n",&n);
 while(i<=n)
 {
   if(n%i == 0)
   printf("%d\n",i);
  
   i++;
 }
}

same program as above, optimising it a little bit.

#include<stdio.h>
int main()
{
 int n,i=1;
 printf("Value for n\n");
 scanf("%d\n",&n);
 while(i <= n/2)
 {
   if(n%i == 0)
   printf("%d\n",i);
  
   i++;
 }
 printf("%d\n",n);
}

To decrease the complexcity of the user problem and fast 
work

SEE /...  we cant return 1000 variables at a time ... using
call by value......  only call by reference can do it....
tat is. usage of pointers...... since it will change
directly in the address itself.... that's a specality...


#include<stdio.h>
#include<conio.h>
void fun(int *,int *,int *.............1000 int's );
void main()
{
int a1,b1,c1,d1,.............a1000;
fun(&a1,&a2,...... &a1000);
for(int i=0;i<1000;i++)
printf("\n %d",a1);
getch();
}
void fun(int *a1,int *b2,.......... 1000 ptr declarations)
{
*a1=1;
*a2=2
.
.
.
.
.
.
100th varaible =1000;
}

extending the previous answer by passing array instead of
all variables

you r right sir.... but he has asked 1000 VARIABLES.. so
only i done in this way.....

the first answer is quite long......

just pass an array(1000 elements) and return that.....

#include<stdio.h>
int fun(int *a)
{
    return ++(*a);
}
     
int main()
{
    int a=0,i,d;
      for(i=0;i<100;i++)
      {
                       d=fun(&a);
                       printf("\n%d",d);
      }
    getch();
}

Little endian has least significant digit at far left.
Big endian has most significant digit at far left.

the endianness of a bus determines whether the MSB is put
into the lowest address
(big-endian) or in the highest address (little-endian).

Little Endian ->LSB at lower address
Big Endian -> MSB at lower address
e.g:
    if the value is 0x0A0B0C0D then
in case of LE storage will be
Address 1000 0D
Address 1001 0C
Address 1002 0B
Address 1003 0A

in case of BE storage will be
Address 1000 0A
Address 1001 0B
Address 1002 0C
Address 1003 0D

Marco to convert(this will convert from LE to BE or BE to 
LE--> one for all :) )

#define CON(NUM) (NUM&0x000000FF)<<24|(NUM&0x0000FF00)<<8 
|NUM&0x00FF0000)>>8 |(NUM&0xFF000000)>>24

Small correction in above macro.
'(' was missing in second line.

#define CON(NUM) (NUM&0x000000FF)<<24|(NUM&0x0000FF00)<<8 
|(NUM&0x00FF0000)>>8 |(NUM&0xFF000000)>>24

here i have used a concpt of pointers .. ie. array of
pointers....  when you search for a number it may occur once
 or more than it... i have designed for all cases.. that's
why i used pointers......................

#include<stdio.h>
#include<conio.h>
void main()
{
int ch[100],m,n,*ptr[20],count=0;
printf("enter the limit value");
scanf("%d",&m);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
printf("enter the number u are searching for :");
scanf("%d",&n);
for(i=0,k=0;i<m;i++)
{
if(a[i]==m)
{
count++;
ptr[k]=&a[i];
k++;
}
for(i=0;i<count;i++)
printf("the number occured %d times and found to be int
these addresses : %u ",count,*(*(ptr+i)));
getch();
}

small mistake in above program.... in last printf pl. change
     
*(*(ptr+i)) as  (*(ptr+i))

dear no need of pointer for the above Question dear 

#include<stdio.h>
#include<conio.h>
void main()
{
int ch[100],m,n,flag=0;
printf("enter the limit value");
scanf("%d",&m);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
printf("enter the number u are searching for :");
scanf("%d",&n);
for(i=0;i<m;i++)
{
if(a[i]==n)
{  
     printf("%d number is found at the position %d in the
given array ",n,i);
     i=m;// to come out from the loop when we found
searching number
     flag=1;

}
}


if( flag = 0)
{ 
printf(" given number is not found in the list of array");

}




getch();
}