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without using arithmatic operator convert an intger variable
x into x+1
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#include<stdio.h>
void main()
{
int no;
int size, i;
printf("ENTER THE NO: ");
scanf("%d",&no);
size = sizeof(int) * 8;
for(i = 0; i < size; i++)
{
if((no & (0x01 << i)) != 0)
{
no = no^(0x01 << i);
}
else
{
no = no |(0x01 << i);
break;
}
}
printf("OUTPUT :%d \n", no);
_getch();
}
| Is This Answer Correct ? | 1 Yes | 0 No |
Hi,
In addition to the Answer#2, which i have posted already,
here i am posting another program which can be used to add
any two nos without using arithmatic operators.
Note: The program written below follows a logic of binary
addition.
#include<stdio.h>
#include<conio.h>
void main()
{
int no1, no2, size, i;
int res = 0, carry = 0, temp1 = 0, temp2 = 0;
size = sizeof(int) * 8;
printf("ENTER THE TWO NOS TO BE ADDED: \n");
scanf("%d %d", &no1, &no2);
for(i = 0; i < size ; i++)
{
temp1 = (no1 & 0x01 << i) ? 1 : 0;
temp2 = (no2 & 0x01 << i) ? 1 : 0;
if((temp1 & temp2) == 1 && (carry == 1))
{
res = res | 0x01 << i;
carry = 1;
}
else if((temp1 & temp2) == 1 && (carry == 0))
{
res = res | 0x00;
carry = 1;
}
else if((temp1 | temp2) == 1 && (carry == 1))
{
res = res | 0x00;
carry = 1;
}
else if((temp1 | temp2) == 1 && (carry == 0))
{
res = res | 0x01 << i;
carry = 0;
}
else if((temp1 | temp2) == 0 && (carry == 1))
{
res = res | 0x01 << i;
carry = 0;
}
else if((temp1 | temp2) == 0 && (carry == 0))
{
res = res | 0x00;
carry = 0;
}
else
{
/*Fatal Error*/
}
}
printf("\nRESULT: %d", res);
_getch();
}
| Is This Answer Correct ? | 1 Yes | 0 No |
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