here the return function will give an error message or it 
will only take the first function (ie) fibbo(n-1) since 
after return this is the first recursive function 
called.... so this altast return 1 to the main program.... 
that's all.... as for as i know this will be the 
procedure...... and then the "or" must not be used .. only 
logicalOR must be used ||.........

#include <stdio.h>

int fibonacci ( int nNumber )
{
	if ( ( nNumber == 0 ) || ( nNumber == 1 ) ) return 
( nNumber );
	return  fibonacci ( nNumber -1 )  + fibonacci ( 
nNumber - 2 ) ;
}


int main ( int argc, char* argv[] )
{
	printf ( "\n The Fibnoci value :%d", fibonacci ( 
5 ) );
	return ( 1 );


Other than the logical or, everyting is perfect, the 
function will recursivel bubble down and for this value it 
ud become like this if u copy this to a notepad, with 
formating, it ud be easy to understand

				4 		+ 	
				3

		3		+		2	
		2		+		1

	2	+	1		1	+	0
	1	+	0		( will return 1 )

1 	+	0 ( all others will return 1 )

using the function 
gets(variable)

#include<stdio.h>
#include<conio.h>
void main()
{
char a[50];
printf("enter the string :");
for(int i=0;1;i++)
{
scanf("%c",&a[i]);
if(a[i]=='\n')
a[i]='\0'
break;
}
for(i=0;a[i]!='\0;i++)
printf("%c",a[i]);
getch();
}

#include<stdio.h>
#include<conio.h>
void main()
{
char a[50];
printf("enter the string:");
scanf("%s",&a);
printf("%s",a);
getch();
}

char a[50];
use scanf(" %[^\n]",a);

#include<stdioi.h>
void main()
{
int n,count=0,a[20];
scanf("%d",&n);
for(i=0;n>0;i++)
{
a[i]=n%10;
n=n/10;
count++;
}
for(i=0;i<count;i++)
{
switch(a[i])
{
case 1:
printf("one");
break;
case 2:
printf("two");
break;
  like this print upto case 0 , (ie) after case 9 ... put
case 0.... and finish the switch and close the loop.......

#include<stdio.h>
#include<conio.h>
void main()
{
int a,b;
printf("Enter the number:");
scanf("%d%d",&a,&b)
a=a+b;
b=a-b;
a=a-b;
printf("%d%d",a,b);
getch();

}

void main()
{
int a=10,b=20;
a^=b^=a^=b;
printf("a=%d,b=%d",a,b)
getch();
}

void main()
{
int a[2]={20,10};
a[0]=a[0]^a[1];
a[1]=a[0]^a[1];
a[0]=a[0]^a[1];
printf("a=%d,b=%d",a[0],a[1])
getch();
}

void main()
{
int a,b;
printf("enter the two numbers\n");
scanf("%d%d",&a,&b);
printf("a=%d\n b=%d\n",a,b);
a=a+b-(b=a);
printf("a=%d\n b=%d\n",a,b);
getch();
}

void main()
{
 int a,len,palin;
 char s[20];
 
 scanf("%s",s);
 
 len=strlen(s);
 printf("%d \n",len);
 
 for(a=0,len=len-1;a<len;a++,len--)
 {
     if(s[a]==s[len])
         palin=1;
      
	  else 
           break;
 }

  if(palin==1)
     printf("string %s is palindrome",s);
 else
      printf("string %s is not palindrome",s);

getch();

}

#include <stdio.h>

int isPalindrome ( char* str, int nLength )
{
	if ( nLength < 1 ) return ( 1 );
	if ( str [0] == str [ nLength -1 ] ) return ( 
isPalindrome ( ( str + 1 ) , ( nLength - 2 ) ) );
	else return ( 0 );
}

int main (int argc, char* argv[])
{
	char a[10] = {"ropepor"};
	if ( isPalindrome  ( a, strlen ( a ) ) ) printf 
("\n The string is Palindrome");
	else printf ("\n The string is NOT Palindrome");
	return (1 );

void main()
{
 int a,len,palin;
 char s[20];
 
 scanf("%s",s);
 
 len=strlen(s);
 printf("%d \n",len);
 
 for(a=0,len=len-1;a<len;a++,len--)
 {
    palin=0;
     if(s[a]==s[len])
         palin=1;
      
	  else
{ 
printf("string %s is not palindrome",s);
getch();
exit();
}
  }
printf("string %s is palindrome",s);    
getch();

}

bool
IsPalindrome( char *lpStr )
{
	int n,i;
	n = strlen( lpStr );
	for ( i = 0; i < n/2; i++ )
	{
		if ( lpStr[i] != lpStr[n-i-1] )
		{
			return FALSE;
		}
	}
	return TRUE;
}

#include"string.h"
void main()
{
char *str,*rev;
int i,j;
clrscr();
printf("\nEnter a string:");
scanf("%s",str);
for(i=strlen(str)-1,j=0;i>=0;i--,j++)
rev[j]=str[i];
rev[j]='\0';
if(strcmp(rev,str))
printf("\nThe string is not a palindrome");
else
printf("\nThe string is a palindrome");
getch();
}

error.
because, there is no header file.
and no ";" is given after the end of classes.

there is no include file iostream for cout
immproper ending for classes ';'

#include<stdio.h>
#include<conio.h>
void main()
{
char a[]="enter the no. of terms tou are going to enter :";
char a1[100];
char n;
puts(a);
int flag=0;
n=getchar();
int n1;
n1=(int)n    /* type casting*/
   for(int i=0;i<n1;i++)
      {
    a[i]=getchar();
}
char a3[]="enter the number do you want to find :";
puts(a3);
char s;
  s=getchar();
int b=(int)s;
  for(i=0;i<n1;i++)
{
     if(b==(int)a[i])
flag=1;
   }
if(flag==1)
printf("number is found");
else 
printf("not found");
getch();
}
this was the logic suddenly striked me.......................

very sorry yaar/.... i forgettenly used printf statements in
my before posts.......... i think this can be the logic...
if am not wrong..............

#include<stdio.h>
#include<conio.h>
void main()
{
char a[]="enter the no. of terms tou are going to enter :";
char a1[100];
char n;
puts(a);
int flag=0;
n=getchar();
int n1;
n1=(int)n    /* type casting*/
   for(int i=0;i<n1;i++)
      {
    a[i]=getchar();
}
char a3[]="enter the number do you want to find :";
puts(a3);
char s;
  s=getchar();
int b=(int)s;
  for(i=0;i<n1;i++)
{
     if(b==(int)a[i])
flag=1;
   }
if(flag==1)
{
char q[]="number is found";
puts(q);
}
else
{
char w[]="number not found";
puts(w);
}
getch();
}

Dear... Vignesh... r u tested?
if tested..., on what Compiler...

ya..... sorry... it wont work... since getchar gets only one
character the time.... when we give 48... i will take 4 and
8 .. but not as 48.... sorry for posting it... a lesson i
have learnt

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

int main ( int argc, char* argv[] )
{
	int array [ 10 ] = { 546, 541, 128, 1027, 1000, 
10234, 657, 343, 111, 272 };
	char _variable [ 50 ];
	int _count = 0;
	int _value = 0;

	memset ( _variable, '\0', 50 );
	puts ("Enter the number to search: ");
	gets ( _variable );
	
	_value = atoi ( _variable );

	for ( _count = 0; _count < 10; _count++ ) {
		if ( array [ _count ] == _value ) {
			puts ("Search SUCCESSFUL");
			return ( 0 );
		}
	}
	puts ("Search NOT SUCCESSFUL");
	return ( 1 );
}

If u guys really want to check the given input is numeric, 
use isnumeric function defined in ctypes.h

sorry for the wrong information use isdigit(char)

#
#
void main()
{
int num,n1,n2,sum;
cout<<"enter a 4 digit no.";
cin>>num;
n1=num/1000;
n2=num%10;
sum=n1+n2;
cout<<"sum of 1st & 4th digit is"<<sum;
}

#include<stdio.h>
main()
{
int n,n1,n4;
printf("Enter 4 digit number:");
scanf("%d",&n);
n4=n%10;
n1=n/100;
printf("\nResult=",(n1+n4));
getch();
}

thiz z how 2 find the sum of first and last digit of any 
number

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
long int n,nf,nl,temp;
int count=-1;
printf("Enter a number:");
scanf("%ld",&n);
temp =n;
while(temp<0)
{
  temp=temp/10;
c++;
}
nl=n%10;
nf=n/(pow(10,c));
printf("\n%ldResult=",(nf+nl));
getch();
}

#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int num,n1,n2,sum;
clrscr();
printf("Enter 4 digit number:");
scanf("%d",&num);
n1=num/1000;
n2=num%10;
sum=n1+n2;
printf("Sum of 1st & 4th digit number is=%d",sum);
getch();
}
enjoy!!!!!!!!!!!!

any program just mail me:
ssmartinp@gmail.com

main()
{
printf("*\n**\n***);
}
OR
main()
{
int n,i;
while(n<3)
{
i=0;
while(i<=n)
{printf("*");i++;}
printf("\n); n++;
}

void main()
{
int i,j;
for(i=0;i<3;i++)
printf("\n");
for(j=0;j<i;j++)
printf("*");
}

# include <Stdio.h>
# include <conio.h>
# include <string.h>

void main()
{
	int j,i,n;

	scanf("%d",&n);

	for (i=0;i<=n;i++)
	{
		for(j=0;j<i;j++)
			printf("*");
		printf("\n");
    }

	getch();

}

#include<stdio.h>
main()
{
int n,i=0,j=0;
printf("Enter an intger : ");//upto n nof *'s
scanf("%d",&n);
while(j<n)
        {
        for(i=0;i<=j;printf("*"),i++);
        printf("\n",j++);
        }
}

void main()
{
int i,j;
for(i=0;i<3;i++){
  for(j=0;j<i;j++){
    printf("*");
    printf("\n");
  }
}
}

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

/**
 * int copy ( char* str, char* _pattern_to_copy, int 
number_of_bytes_moved, int total_size_allocated, int 
diff_size )
 * @param str, the pointer from where the pattern starts
 * @param _pattern_to_copy, what needs to be placed insted 
of the pattern
 * @param number_of_bytes_moved, the number of bytes at 
which the pattern first appeared
 * @param total_size_allocated, the total size the space 
has been allocated (eliminating the space for NULL)
 * @param diff_size, the difference between the pattern and 
the string that needs to be replaced
 *
 */
int copy ( char* str, char* _pattern_to_replace, int 
number_of_bytes_moved, int total_size_allocated, int 
diff_size )
{
	int _loop = 0;

	/**
	 * Starting from end, move towards untill the 
pattern appears
	 * start point is array start + the 
number_of_bytes_moved (first apperance of the pattern)
	 * so the end point should be array end - 
number_of_bytes_moved (the last space before the NULL 
character
	 * move the array down to create space for the 
string to be replaced 
	 */
	for ( _loop = total_size_allocated; ( 
total_size_allocated - number_of_bytes_moved ) >  
diff_size; total_size_allocated-- )
	{
		str [ total_size_allocated - 
number_of_bytes_moved ] = str [ total_size_allocated - 
number_of_bytes_moved - diff_size ];
	}

	/**
	 * Replace the string on the pattern
	 * Use memcpy since it does not copy provide NULL 
by itself
	 */
	memcpy ( str, _pattern_to_replace, strlen ( 
_pattern_to_replace ) );

	return ( 0 );
}

int main ( int argc, char* argv [] )
{
	char* string_value = NULL;
	char pattern_to_replace [] = {"replacement"};
	char pattern_2b_replaced [] = {"#"};
	int _count = 0;
	char* _ptr = NULL;
	int alocation_size = 0;
	int number_of_bytes_moved = 0;
	int old_strlen = 0;

	/**
	 * Allocate the size to hold the orignal string
	 * copy the required string with the pattern
	 */
	string_value = ( char* ) malloc ( 26 * sizeof ( 
char ) );
	if ( NULL != string_value ) 
	{
		strcpy ( string_value, "Hi # of new # 
string by #" );
	}

	/**
	 * Count the number of occurences of the pattern.
	 */
	_ptr = string_value;
	while ( _ptr = strstr ( _ptr, 
pattern_2b_replaced ) )
	{
		_ptr += 1;
		_count++;
	}

	/**
	 * Calculate the size of the new string that needs 
to be accomadated
	 * the number of times the pattern occurs in the 
original string
	 */
	alocation_size = ( strlen ( pattern_to_replace ) - 
strlen ( pattern_2b_replaced ) ) * _count; // calculated 
size of new string
	alocation_size += strlen ( string_value ); 	
			          // add the size of old 
string
	old_strlen = strlen ( string_value ) + 1;	
				  // allocate space to 
store NULL

	/**
	 * Increase the space in the old pointer using 
realloc
	 * and copy '&' in the newly allocated spaces alone
	 * just for the sake of our reference.
	 *
	 * Remember to exclude the 1 which we allocated for 
the NULL character
	 */
	string_value = ( char* ) realloc ( string_value, ( 
alocation_size * sizeof ( char ) ) );
	memset ( ( string_value + old_strlen) , '&', ( 
alocation_size - old_strlen - 1 ) );

	/**
	 * Store the NULL at the end of the total allocated 
size
	 * we use alocation_size - 1
	 * 
	 * eg: a[5] means 6th location in the array
	 * If we want to store at the 6th location 
	 * we need to provide 5 (ie. alocation_size - 1)
	 */
	string_value [ alocation_size - 1 ] = '\0';

	/**
	 * Replace the string needs to be filled
	 */
	_ptr = string_value;
	while ( _ptr = strstr ( _ptr, 
pattern_2b_replaced ) )
	{
		number_of_bytes_moved = ( _ptr - 
string_value );
		copy ( _ptr, pattern_to_replace, 
number_of_bytes_moved, ( alocation_size - 1 ), 
				( strlen ( 
pattern_to_replace ) - strlen ( pattern_2b_replaced ) ) );
		_ptr += 1;
	}


	printf ( "\n string :%s, no of occurence :%d, 
size :%d", string_value, _count, alocation_size );

	free ( string_value );
}

no this answer is not correct

fail
because here if is taking 0 boolean value means condition
becomes false and else will be executed

fail

FAIL



THANK U

C as a language is portable across any platforms, as it 
needs a compiler to compile in different platforms.

The drawback of compiled 'C' code is "Architecture 
Dependent" and hence the code compiled in 32 bit 
architecture cannot be used for 64 bit.

#include<stdio.h>
main()
{
   if(Printf("Here is the message without terminator"))
   {
   }
}

we can make without using if,for,while statments.....

main()
{
  fun(printf("\nCute Ramya !!!!!"));
}
fun(int i)
{
;;;;; i++
}

#include<stdio.h>

void main()
{
  fun(printf("\nCute Ramya !!!!!"));
}
fun(int i)
{
i++;
}

#include<stdio.h>
#include<conio.h>
main()
{
      if("adesh")
       {
        }
}

#include<stdio.h>
main()
{
 if(printf("the message is print without terminator"))
}

c is a procedural language and C++ is object oriented 
language

c uses the top-down approach.c++ uses bottom-up approach

c having no security for data..but c++ having that 
security,using class(access specifier).

in C structures are used...but in cpp classes are 
used..which is more secured