adding and subtracting a long int binary number in c - c

Question

I have a long int number which is in binary form.
I want to subtract or add another binary number from this number keeping the data type of destination variable same, i.e long int
How can I do it?
I had a solution in which destination variable was an array of int.
what I tried so far
int main() {
long int binary1;
long int binary2 = 0001;
int i = 0, rem = 0;
int sum[20];
printf("Enter binary number");
scanf("%ld", &binary1);
while (binary1 != 0 || binary2 != 0) {
sum[i++] =(binary1 % 10 + binary2 % 10 + rem) % 2;
rem =(binary1 % 10 + binary2 % 10 + rem) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (rem != 0)
sum[i++] = rem;
--i;
printf("Sum of two binary numbers: ");
while (i >= 0)
printf("%d", sum[i--]);
return 0;
}
Here the result is stored as int. I want to store it in long int because I need to use this sum for the next addition operation.
How can I do the next addition operation where sum is one of the operand? sum here is in int, and another operand is in long int.

Answer 1

Store as chars it what you want to do I think
#include <stdio.h>
#include <stdlib.h>
#include <strings.h>
#include <string.h>
int main() {
long int binary1;
long int binary2 = 0001;
int i = 0, rem = 0;
char sum[20];
printf("Enter binary number: ");
scanf("%ld", &binary1);
while (binary1 != 0 || binary2 != 0) {
sum[i++]=(binary1 % 10 + binary2 % 10 + rem) % 2 > 0 ? '1' : '0';
rem =(binary1 % 10 + binary2 % 10 + rem) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (rem != 0)
sum[i++] = rem;
--i;
printf("Sum of two binary numbers: ");
while (i >= 0)
printf("%c", sum[i--]);
printf("\n");
return 0;
}

Answer 2

your problem is that
long int binary1=1011;
will load your variable with 1011 dec which is 1111110011 bin you can use standard addition operator but different load/print operations. If you want to stick to your DCB (decimal coded binary) encoding (similar to BCD binary coded decimal) then you need to rewrite all math operation. I think simpler would be this:
long int bin2dec(long int bin) // DCB -> binary
{
long int x,m;
for (m=1;m<bin;m*=10);
for (x=0;m;m/=10)
{
x<<=1;
x+=bin/m;
bin%=m;
}
return x;
}
long int dec2bin(long int dec) // binary -> DCB
{
long int x,m;
for (m=1;m<dec;m<<=1);
for (x=0;m;m>>=1)
{
x*=10;
x+=dec/m;
dec&=m-1;
}
return x;
}
void main()
{
long int b1=1011;
long int b2= 1;
long int b3= 100;
long int sum;
sum = dec2bin(bin2dec(b1)+bin2dec(b2)+bin2dec(b3));
// here print the sum or whatever
}
The point is to convert DCB into standard binary representation before math operations and then convert the result back to DCB for printing.