C++ Adding Two 16 bit binary numbers together

Question

I am trying to add two 16 digit binary numbers together. Each binary number will have been entered by the user and saved as: int binaryOne[16] and int binaryTwo[16]. Originally,

int binaryOne[16] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int binaryTwo[16] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

are the values declared. Then, the user has the option to change these values using buttons that I implement before the Solution() function. So, once the used gets to the Solution() function, the numbers may be stored as:

int binaryOne[16] = {0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1};
int binaryTwo[16] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1};

This is what I have so far:

void Solution() {
  /*
    Rule of binary addition:
    0 + 0 = 0
    1 + 0 = 1
    0 + 1 = 1
    1 + 1 = 1 and carry = 1
   */
   lcd.clear();
   lcd.setCursor(0,0);
   lcd.print("Solution");
   lcd.setCursor(0,2);





    int a[16] = binaryOne[16];
    int b[16] = binaryTwo[16];
    int i = 0;
    int remainder = 0;
    int sum[32];

    while(a[16] != 0 || b[16] != 0){
      sum[i++] = (a[16] % 10 + b[16] % 10 + remainder) % 2;
      remainder = (a[16] % 10 + b[16] % 10 + remainder) / 2; 
      a[16] = a[16] /10;
      b[16] = b[16] / 10;
    }
    if(remainder != 0){
      sum[i++] = remainder;
    }
    i--;
    while(i>=0){
      lcd.print("%d", sum[i--]);
   }
   return 0;
}

I believe I am doing something wrong with calling the arrays since I am renaming them to a[16] and b[16] within this function. I'm also curious, since I only have 16 bits of space on the screen, what if the user enters a binary number such as 1111111111111111 + 1000000000000000 and therefor there is going to be a carry on that first digit? Would it be best just to have an error message?

score 2 · Accepted Answer · answered Apr 05 '18 at 18:11

You're making a mountain out of a molehill.

You're trying to manually manipulate binary digits in an array, but C can already manipulate binary digits. Everything is binary already - you just need to represent it as such.

All you need to do is convert the contents of your array into an actual number, then do simple mathematics on that number. One of the simplest ways of doing that is to shift the bits into a variable:

For instance:

int binaryOne[16] = {0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1};
int binaryTwo[16] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1};

uint16_t binOne = 0;
uint16_t binTwo = 0;

for (int i = 0; i < 16; i++) {
    binOne <<= 1;
    binOne |= binaryOne[i];
    binTwo <<= 1;
    binTwo |= binaryTwo[i];
}

Now you can add them together, since they're just numbers:

uint32_t result = binOne + binTwo;

And you can output it as binary:

Serial.println(result, BIN);

If you want to convert back into an array again (why?!) you can just do the opposite as before:

bool resultBin[32];

for (int i = 0; i < 32; i++) {
    resultBin[i] = (result & (1 << (31-i))) ? 1 : 0;
}

score 2 · Answer 2 · answered Apr 05 '18 at 18:21

A few issues here:

The way you “rename” the arrays is wrong, and furthermore there is no point in doing such renaming.
There is no reason to divide by 10 (ten, in decimal) within your code: as everything you do is binary, the number ten has nothing to do with what are trying to accomplish.
While doing the addition, you forgot to loop over the bits.
You shouldn't return 0 in a function of type void. In fact, you shouldn't return any value.

Fixing those issues should give you something similar to this:

void Solution() {
    lcd.clear();
    lcd.setCursor(0, 0);
    lcd.print("Solution");
    lcd.setCursor(0, 2);

    // Compute the sum.
    int sum[16];
    int remainder = 0;
    for (int i = 0; i < 16; i++) {
        int s = binaryOne[i] + binaryTwo[i] + remainder;
        sum[i] = s % 2;
        remainder = s / 2;
    }

    // Print it.
    for (int i = 15; i >= 0; i--) {
        lcd.print("%d", sum[i]);
    }
}

Note that there is no point in computing a 32-bit result if you can only print 16 bits: just accept that the computation is done modulo 2¹⁶, just like a 16-bit ALU would do, or the unsigned int data type in the Arduino.

Of course, the whole idea of doing the computation one bit at a time is highly inefficient. See Majenko's answer for a more idiomatic and efficient way of managing numbers bit-wise in C and C++.

C++ Adding Two 16 bit binary numbers together

2 Answers2