adding two binary numbers(integer array) - c++

Question

I have done this program to add two binary numbers of the same length stored in an integer array and store the sum in a new integer array. But due to some logical error, it does not show the desired output.
#include <conio.h>
#include <iostream.h>
void main() {
clrscr();
int a[] = {1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1,
1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1};
int b[] = {1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1,
1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0};
int temp[100];
int carry = 0, p = 35;
for(int i = 34; i >= 0; i--) {
if(a[i] + b[i] + carry == 0) {
temp[p] = 0;
carry = 0;
p--;
}
if((a[i] + b[i] + carry) == 1) {
temp[p] = 1;
carry = 0;
p--;
}
if((a[i] + b[i] + carry) == 2) {
temp[p] = 0;
carry = 1;
p--;
}
if((a[i] + b[i] + carry) > 2) {
temp[p] = 1;
carry = 1;
p--;
}
}
for(int pop = 0; pop < 36; pop++) cout << temp[pop];
getch();
}
the expected output is:
110100101001100101101010110101011111
actual output is:
101100110011011101011011101011011111

Answer 1

if((a[i] + b[i] + carry) == 2) {
temp[p] = 0;
carry = 1;
p--;
}
if((a[i] + b[i] + carry) > 2) {
temp[p] = 1;
carry = 1;
p--;
}
Go through this with a[i] = 1, b[i] = 1, carry = 0
This will trigger the first if here as it should, which then sets carry to 1.
Then the second if will find the configuration a[i] = 1, b[i] = 1, carry = 1 which triggers it despite the original configuration not fitting, causing the bug.
For the future, please get to know how to use a debugger, or at least how to include some debugging console prints in the program. Set stop points or prints where there would be nodes in a corresponding diagram to see if the flow takes the right route.
With prints:
for(int i = 34; i >= 0; i--) {
cout << "entered loop" << endl;
if(a[i] + b[i] + carry == 0) {
temp[p] = 0;
carry = 0;
p--;
cout << "recognized case 1" << endl;
}
if((a[i] + b[i] + carry) == 1) {
temp[p] = 1;
carry = 0;
p--;
cout << "recognized case 2" << endl;
}
if((a[i] + b[i] + carry) == 2) {
temp[p] = 0;
carry = 1;
p--;
cout << "recognized case 3" << endl;
}
if((a[i] + b[i] + carry) > 2) {
temp[p] = 1;
carry = 1;
p--;
cout << "recognized case 4" << endl;
}
}
(But using the debugger is better than prints.)

Answer 2

The issue in your code is you have a set of separate if statements which each modify carry, if carry is changed in such a way to alter the result of the subsequent if statements then multiple if statement bodies will be executed.
The simple fix is to change your last 3 ifs to else if.
You also need to add the final carry bit to your output:
temp[0] = carry;
for(int pop = 0; pop < 36; pop++) cout << temp[pop];
However your code could be greatly simplified using a swtich statement. Simpler code is generally less buggy and easier to understand:
for(int i = 34; i >= 0; i--, p--) {
switch(a[i] + b[i] + carry) {
case 0:
temp[p] = 0;
carry = 0;
break;
case 1:
temp[p] = 1;
carry = 0;
break;
case 2:
temp[p] = 0;
carry = 1;
break;
case 3:
temp[p] = 1;
carry = 1;
break;
default:
// should never be reached with inputs of 0 or 1
throw std::invalid_argument("invalid input");
}
}
This code could be simplified even further using bit manipulations to remove all branches and set the bits directly but I'll leave that as an exercise for the reader hint.