what is 3rd venturing graphics...

#include <stdio.h>
      int add(int a, int b)
      {
	    if (!a)
		return b;
	    else
		return add((a & b) << 1, a ^ b);
      }

      int main()
      {
	    unsigned int a,b;
	    printf("Enter the two numbers: \n");

	    scanf("%d",&a);
	    scanf("%d",&b);
	    printf("Sum is: %d",add(a,b));
      }

Please explain me the code

Hi,
This is the code for a FULL ADDER circuit.

why cant we just or the two numbers

Hi Nitish,

If we will do the LOGICAL OR function then we will get
either 1 or 0. 
If we will do the BITWISE OR then we will get the largest of
the two.. 

For
 ex:
Let us take, First number as 2 and Second number as 3..
Then as per the first case we will get 1 as the output. 
10(2) || 11(3) -> 1(1)
As per the second case the output will be 3..
10(2) | 11(3) -> 11(3).
Ok

int sum(int num1,int num2)
{	
     for(int i=0;i<num2;i++)	
        num1++;	
     return num1;
}

even this gives the same ans as the above program gives...
just every one plz check it and tell me
#include<stdio.h>
#include<math.h>
int main()
{
int a,b,c;
printf("enter the nos");
scanf("%d %d",&b,&c);
a= (b^c);
printf("%d",a);
return 0;
}

#include<stdio.h>
int a,b,c;
{
printf("enter the two values");
scanf("%d","%d",&a,&b);
c=a||b;
}

prashant answer is wrong suppose add two similar numbers 
prashant answers will fail because addtion of two similar 
bits  according to the bitwise xor fails.

#include<stdio.h>
main()
{
int a=5,b=4,c;
c=a||b;
printf("sum="+c);
}

# include <stdio.h>
main()
{
int a=8,b=2;
a|=b;
printf("sum="+a);
}

#include <stdio.h>

char fun ( char a, char b, int flag )
{
	if ( flag ) return ( a |= ( flag << ( (int) b - 
1 ) ) );
	return ( a &= ~( 1 << ( (int) b - 1 ) ) );
}

int main ( int argc, char* argv [] )
{

	char a = 45;

	printf ( "\n Before change :%d", (int) a );
	printf ( "\n After change :%d", (int) fun ( a, 
(char) 7, 1 ) );

	return ( 0 );
}

18) d 
o/p is 
2 5
19)
d
o/p is
100 200 100
20) none
Answer is 1,2 & 4.

After calling the function swap(), the values of x,y will be
the same.

i.e. x = 2, y = 3.

The scope of the variables x,y,temp in the swap() function
lies inside the function swap() itself. So there will not be
any change in the values of x,y in the main() function..

the values will be x =2 and y = 3.

the variables x and y declared in main() are local to main.

whereas 

the variables x and y declared in swap() are local to swap..

 the change in the value of the variables in either 
function will have zero effect on the other function.

Hence the value remains teh same.

d

c

a

b

ans is C

d

a

Answer is (d).

c

Ans is (c)

c

D

ans is c....write

a

The Output will be:

First output  : 12
Second output : 13
Third output  : 14

for changevalue() function:
      Even though the value of x(11) is sent to
changevalue() and it returns x(12), no variable is assigned
to capture that 12. So, in main() function, x remains as 11(
not as 12 ) . then it gets incremented and prints the value
12...

And, the Same story for other functions also.....

The Output will be:

First output  : 12
Second output : 13
Third output  : 13

Static is protected one and you cannot acess the variable 
in other files/functions.It is declared in the top of the 
file/function.The value retains tho,t the program.

Static remains throughout the program, however its scope is 
limited to that file alone. If a program consists of 
multiple files, you want a variable to be seen by the 
entire file and do not want that variable to be seen by the 
other files, then mark is as static

static int a; 

as a global variable

#include<stdio.h>
#include <conio.h>
#include<string.h>
#include<alloc.h>
char * mul(char *, char);
char * add(char *, char *);
void main()
{

	char *no1=(char * )malloc(100);
	char *no2=(char *)malloc(100);
	char *q="0";
	char *p;
	int i=0,j=0,t=0,l=0;
	clrscr();
	printf("\n Enter the large no ");
	gets(no1);
	printf("\n Enter the second no ");
	gets(no2);

	while(l<strlen(no2))
	{
		p=mul(no1,no2[l]);
		for(j=1;j<=strlen(no2)-l-1;j++)
		strcat(p,"0");
		q=add(q,p);
		l++;
	}
	printf("\n Multiplication is %s",q);
	free(no1);
	free(no2);
}


char * mul(char *x, char ch)
{
	int i=0,j=0,t=0;
	char *p=(char *)malloc(300);
	char *a =(char *)malloc(300);
	strcpy(p,x);
	strrev(p);
	while(p[i]!='\0')
	{
		a[j]=(p[i]-48)*(ch-48)+t;
		t=a[j]/10;
		a[j]=(a[j]%10)+48;
		i++;
		j++;
	}
	if(t!=0)
	a[j]=t+48;
	else
	j--;
	a[j+1]='\0';
	strrev(a);
	free(p);
	return(a);
}
char * add(char *p, char *q)
{
	char *t=(char *)malloc(300);
	int i=0,j=0,x=0,a;
	strrev(p);
	strrev(q);
	while(p[i]!='\0' && q[i]!='\0')
	{
		a=(p[i]-48)+(q[i]-48)+x;
		x=a/10;
		a=a%10;
		t[i]=a+48;
		i++;
	}
	while(i<strlen(p))
	{
		a=(p[i]-48)+x;
		x=a/10;
		a=a%10;
		t[i]=a+48;
		i++;
	}
	while(i<strlen(q))
	{
		a=(q[i]-48)+x;
		x=a/10;
		a=a%10;
		t[i]=a+48;
		i++;
	}

	if(x!=0)
		t[i++]=x+48;
	t[i]='\0';
	strrev(t);
	return(t);
}

i++* is meaningless , do u want to ask *++i and *i++ diff ?

*++i -->  it increments the i then access the value poiting 
by i

*i++ --> it first access the value pointed by i , then 
increment the i ( increments pointer , not value)

The postfix ++ and -- operators essentially have higher
precedence than the prefix unary operators. Therefore, *i++
is equivalent to *(i++); it increments i, and returns the
value which i pointed to before i was incremented. To
increment the value pointed to by i, use (*i)++ (or perhaps
++*i, if the evaluation order of the side effect doesn't
matter).

Ref:comp.lang.c FAQ list · Question 4.3

i++* wont work .... as for as i know.... it's meaningless

comin to *++i, i is a pointer holding an address so here ++ 
and * holds the same priority so we ll go for associativity 
of these operators. it's RIGHT to LEFT.

so , address in 'i' will get incremented and then if that 
address points to some value means it will print that value 
or else it will have garbage value



thank u

can be done by reversing the bits

#include<stdio.h>
#include<conio.h>

int main()
{
	int i,number,count=0,a[100];

	printf("Enter the number\n");
	scanf("%d",&number);

	for(i=7;i>=0;i--)
	{
        if((1<<i) & number)
			a[count] = 1;
		else
			a[count] = 0;
		
		count++;
	}

	printf("Binary Value of the Given Number is:\n");
	for(i=0;i<=7;i++)
	{
		printf("%d",a[i]);
	}
	printf("\nReversed Binary Value of the Given Number is:\n");
	for(i=0;i<=7;i++)
	{
		printf("%d",a[7-i]);
	}
	printf("\n");
}

You can swap swap bits using two ways
1 ) with for loop
2 ) using recursion

#include <stdio.h>

char swap_bits_in_byte ( unsigned char byte_2_swap )
{
	unsigned char swapped_byte = 0;
	int nloop = 0;
	for( nloop = 0; nloop < 8; ++nloop ) { 
		swapped_byte = swapped_byte << 1;
		swapped_byte |= ( byte_2_swap & 1 );
		byte_2_swap = byte_2_swap >> 1;
	}
	return ( swapped_byte );
}


unsigned char swap_bits ( unsigned char byte_2_swap, int 
n_size )
{
	unsigned char swapped_byte = 0;
	int bits = ( ( sizeof ( unsigned char ) * 8 ) - 1 );

	if ( bits == n_size ) {
		swapped_byte = ( byte_2_swap 
				& (unsigned char) ( pow ( 
2, n_size ) ) ) 
			? ( 1 << ( bits - n_size ) ) : 0;
	} else {
		swapped_byte = swap_bits ( byte_2_swap, 
n_size + 1 );
		swapped_byte |= ( byte_2_swap 
				& (unsigned char) ( pow ( 
2, n_size ) ) ) 
			? ( 1 << ( bits - n_size ) ) : 0;
	}
	return ( swapped_byte );
}

int main ( int argc, char* argv [] )
{
	unsigned char byte = 128 | 32;
	unsigned char swapped_byte = 0;

	swapped_byte = swap_bits_in_byte ( byte );
	printf ( "\n Un Swapped Byte :%d", byte );
	printf ( "\n Swapped Byte :%d", swapped_byte );

	swapped_byte = swap_bits ( byte, 0 );
	printf ( "\n Un Swapped Byte :%d", byte );
	printf ( "\n Swapped Byte :%d", swapped_byte );

}