main()
{
	char name1[20],name2[20];
	int a,b,i,j,c;
	clrscr();
	printf("Enter the first name:\t");
	gets(name1);
	printf("Enter the second name:\t");
	gets(name2);
	a=strlen(name1);
	b=strlen(name2);

	for(i=0;i<=a;i++)
	{
		for(j=0;j<=b;j++)
		{
			if(name1[i]==name2[j])
			{
				a--;
				b--;
			}
		}
	}
	c=a+b;
	i=c%6;
	switch(i)
	{
		case 0:printf("Friends");
			break;
		case 1:printf("lovers");
			break;
		case 2:printf("Ansisters");
			break;
		case 3:printf("marriage");
			break;
		case 4:printf("enemies");
			break;
		case5:printf("sisters");
			break;
	}

	getch();
}

Answer is :64

++i * ++i * **i
->
2       3     4
now started this way
<-
4   *   4  *   4
=64

The precedence of the operations, should be (reading from
left to right in the equation)

++i   <first ++i i=2>
++i   <second ++i i=3>
*     <first product yields 3*3=9>
++i   <third ++i i=4>
*     <giving the second product 3*4=36>

Thus, the first product (*) is computed before the third ++i
is computed.  Once the first product is completed, i is
incremented to i=4 and the second product can occur now.


Now, if you add some parentheses to the expression giving

++i * (++i * ++i)

then you will get 64, as the other replies suggest.  Tracing
through the order of operations in this one

++i <first ++i i=2> 
++i <second ++I i=3>
++i <third ++I i=4>
*   <the product in the parentheses now yields 4*4=16>
*   <the first * yields 4*16=64>

Here, the first product (*) cannot occur until it knows the
result of the product in the parenthesis.  Thus, all three
increments must occur before the multiplications take place.

to convert it into ascii code just assign the entered 
variable to a int(integer)variabe e.g.
int i=a
which on printing i will gie 64 as the o/p

include<stdio.h> means we include standard input and output 
functions code we does not write any thing about library 
functions .h means header file if we include tis header 
file then we place # before the include<stdio.h> this is c 
syntax.

first create the list of unknown length..... then get the position of the element to be deleted from the user.... the start travelling in the list.... if it encounters the position prescribed by the user ... get the addresses in the list and shift that to the previous node and free this node........


thank u

ya it's possible ..... we can pass whole array as an value....

let's take the code :

void function(char [] );
void main()
{
char ch[30];
function(ch);
getch();
}

void function(char ch[])
{
printf("%s",ch);
}

thank u

hope this will work.....

const char *a : means the string is constant and the pointer
is not...

const char *a="HELLO WORLD" , if we take this example for
the whole scope of the program the string is constant and we
can't assign any other string to that pointer 'a'....

char * const a : means the pointer is constant (address) but
string is not...... 

char * const a="hello world" , if we take this example ,
here the address will be always constant.... string can vary..

char const *a : means string is a constant and pointer is
not..... as we have seen from the first example...


thank u

surely there is some difference.....

here 'a' is represented as array in which string gets stored
in consecutive locations......

p is a pointer variable where string is initilized... so in
p the base address of "literal " will get stored...... 

*p++ increments 'p' , but pertaining to some conditions.....
++ has more precedence than * , so first it will increment
the address and correspondingly it will show the value as *
precedes..... so after the increment the p points to 'i'...



thank u

#include<stdio.h>
void powerOfTwo(int number)
{
 if(!(number & number-1) && number)
     printf("\nthe number is a power of 2\n");
 else printf("\nThe number is not a power of 2\n");
}
 

int main()
{
 powerOfTwo(32); //power of 2
 powerOfTwo(22);  //not a power of 2
 return 0;
}

int main(s){
    s="int main(s){s=%c%s%c;printf(s,34,s,34);return 0;}";
    printf(s,34,s,34);
    return 0;
}

#include<stdio.h>
#include<conio.h>
void main()
{
unsigned i;
int j=0,count=0;;
printf("Enter the number :");
scanf("%ld",&i);
while(j<=31)
{
  if(!(((i>>j)&1)^1))
   count++;
j++;
}
printf("\nnumber of 1's in ur number is : %d",count);
getch();
}


thank u

sir, i can not give the answer of this question.

int main()
{
  if(printf("hello world"))
}

For if statement compiler does not expect semi colon.

int main(){
    if(printf("Hello World")){}
}

#include<stdio.h>
#include<conio.h>
void main()
{
int factorial(int);
int n;
clrscr();
printf("Enter a number: ");
scanf("%d",&n);
printf("Factorial of %d is: %d",n,factorial(n));
getch();
}
int factorial(int f)
{
int fact;
if(f==1)
 return(1);
else
 fact=f*factorial(f-1);
return(fact);
}

the best way what i choose is that : if x=89 , y=-88

x^=y^=x^=y;

this line will swap the above numbers......


thank u

#include<stdio.h>
#include<conio.h>
void main()
{
int a,b;
clrscr();
printf("\n ENTER 2 VALUES : ");
scanf("%d%d",&a,&b);
printf("\n THE VALUES BEFORE SORTING : %d,%d",a,b);
a=a+b;
b=a-b;
a=a-b;
printf("\n THE VALUES AFTER SORTING : %d,%d",a,b);
getch();
}