Add two numbers without using the addition operator | 5 methods

Given two numbers, add them without using an addition operator.

1. Using subtraction operator

int add(int a, int b) {

return a-(-b);

}

2. Repeated Addition/Subtraction using `--/++` operator

#include <iostream>

using namespace std;

int add(int a, int b)

{

// to handle positive `a`

while (a > 0)

{

b++;

a--;

}

// to handle negative `a`

while (a < 0)

{

b--;

a++;

}

return b;

}

int main()

{

cout << add(5, 8) << " ";

cout << add(5, -8) << " ";

cout << add(-5, 8) << " ";

return 0;

}

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Output:

13 -3 3

3. Using `printf()` function

This method makes use of two facts:

We can use an asterisk * to pass the width precision to printf(), rather than hard-coding it into the format string.
printf() function returns the total number of characters printed on the output stream.

Note that we can also use %*c, ' ' replacing %*s, "".

int add(int a, int b)

{

// `%*s` means print a character `*` number of times

return printf("%*s%*s", a, "", b, "");

}

4. Half adder logic

int add(int a, int b)

{

if (!b) {

return a;

}

int sum = a ^ b;

int carry = (a & b) << 1;

return add(sum, carry);

}

5. Using logarithm and exponential function

#include <stdio.h>

int main(void)

{

int a = 8, b = 6;

printf("%g\n", log(exp(a) * exp(b)));

return 0;

}

Add two numbers without using the addition operator | 5 methods

1. Using subtraction operator

2. Repeated Addition/Subtraction using --/++ operator

3. Using printf() function

4. Half adder logic

5. Using logarithm and exponential function

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2. Repeated Addition/Subtraction using `--/++` operator

3. Using `printf()` function