void main()
{
int a = 5;
int b = 7;
int c = 2;
int res;

res = (int)(a/b)?a:b;
res = (int)(res/c)?res:c;

printf("big num = %d",res);
}

#include<stdio.h>
#include<conio.h>
void main()
int a,b,c;
clrscr();
printf("enter any three no.s");
scanf("%d%d%d",&a,&b&c);
if(a>b&&a>c);
{
printf("a is biggest");
}
else if(b>a&&b>c)
{
printf("b is biggest");
}
else
printf("c is biggest");
getch();
}

i need answer and explation to env function

It’s interesting.
This program reads environmental variables. It is as same as
"env" command in unix.
But program is not proper. i.e. if you pass n arguments to
program, then it reads n environmental variables only.
Try this program
void main(int argc,char *argv[],char *env[])
{
int i;
if (2 <=argc){
for(i=0;i<atoi(argv[1]);i++)
printf("\n%s",env[i]);
}else printf("\nPlease enter no. of env variables you want
e.g. 'a.out 5'\n");
}

#define subscript 5


void spiral()
{
    int a[subscript][subscript],i,j,k,l,m,p,q;
    p=q=subscript ;
    for(i=0;i<p;i++,p--)
       {
         for(j=i;j<n-1;j++)
             printf("%d",a[i][j]);
         for(k=i;k<j;k++)
             printf("%d",a[k][j]);
         for(l=k;l>i;l--)
             printf("%d",a[l][k]);
         for(m=k;m>i;m--)
             printf("%d",a[i][m]);
       }
    if(q%2!=0)
       printf("%d",a[j][j]);
}

The main function returns a void value .

The main function does not return any information to the
operating system.

The meaning of void is nothing thats why when the function
cant return any value than at that time we declair it as a void 
void main means main does not return any value

void mean no return type. and main does not return any value so we use void main .

void main means does not  return valu at exicution time 
mens return (0)

void means simply it's return zero(0)

by default main return int to its environment but when we
declare void main it means it will not return any value

void main() is a main function in c language.void means 
nothing return any value.this function is used to execute our 
program.without main(), program can compile but not run.

the below program is for getting two numbers as input
(decimal format) and then the below program will convert in
binary and add WITHOUT ANY ARITHMETIC OPERATORS.....



#include<stdio.h>
#include<conio.h>
#define HIGH 1
#define LOW 0
void main()
{
long c[32],i,n,a,b,k,m,A,CARRY=0;
clrscr();
n=31;
printf("enter the value of a&b:");
scanf("%ld%ld",&a,&b);
for(i=0;i<32;i++)
{
  k=((a>>i)&1);
  m=((b>>i)&1);
   if(!(CARRY^HIGH))
   {
    c[n]=((CARRY^k)^m);
      if(!(k^HIGH)||!(m^HIGH))
       CARRY=1;
      else
       CARRY=0;
   }
   else if(!(k^HIGH) && !(m^HIGH))
   {
    CARRY=1;
    c[n]=k^m;
   }
   else if(!(k^LOW)||!(m^LOW))
   {
    if(!(CARRY^HIGH))
    {
     c[n]=((CARRY^k)^m);
     CARRY=0;
    }
    else
    c[n]=k^m;
   }
  n--;
}
for(i=0;i<32;i++)
printf("%d",c[i]);
getch();
}


thank u

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>
#include<conio.h>
void main()
{
char c,d;
int a;
clrscr();
printf("enter character");
scanf("%c",&c);
a=c;
d=a;
printf("the ascii code is:%d",a);
printf("the character is:%c",d);
getch();
}

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.

actually, in C, every keyword has some specific value some 
specific meaning and those meanings of keywords are already 
been stored in header files like 
<stdio.h>,<conio.h>,<math.h> etc.... so to make computer 
understood the meaning of printf, scanf like keywords, we 
have to use header files in begining of the prog... and # 
is a preprocessor.... this tells computer that now u be 
ready to write a prog... 
this is wat i think.... though i dnt hav technical 
knowledge still i tried to xplain.....

it is library file access,which contains information that 
must be included in the program when it is compiled.

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

I am totally satisfied with your above explanation except
last one. 
i.e.  Char *p="literal";
So, i want to mention yes this will work.
Explanation: *p++. 
Here we have post increment. 
Postfix increment/decrement have high precedence, but the
actual increment or decrement of the operand is delayed (to
be accomplished sometime before the statement completes
execution).
value of printf("\nstr=%c\n",*p++) will be 'l', but before
complete execution of this statement p will point to string
"iteral" as p got incremented.

#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;
}

main()
{
        int i;
        printf("Enter Number :");
        scanf("%d",&i);
        if(i&(i-1))
                printf("Not atwo power");
        else
                printf("Two 's Power");
}

main()
{
        int i;
        printf("Enter Number :");
        scanf("%d",&i);
        if(i&(i-1))
                printf("Not atwo power");
        else
                printf("Two 's Power");
}