Sum of Two Integers Without Using Plus or Minus

The "Sum of Two Integers" problem is a fascinating exercise in bit manipulation, where the objective is to calculate the sum of two integers without using the + or - operators.

Problem Statement

Given two integers a and b, the challenge is to compute their sum using only bitwise operators and without the conventional arithmetic operators.

JavaScript Solution

Here's a JavaScript function that demonstrates the solution:

Breaking Down the Solution

• Carry Calculation: Using let carry = (a & b) << 1, the carry from each bit is computed. The bitwise AND (&) identifies the bits that are 1 in both a and b. The left shift (<< 1) then moves these bits to a higher order, representing the carry effect in binary addition.

• Sum Without Carry: The expression a = a ^ b calculates the sum of a and b without considering the carry. The XOR operation (^) effectively adds the bits where there is at least one 1, aligning with how addition works in binary.

• Iterative Process: A while loop continues this process of updating a with the sum and b with the new carry. This loop iterates until there is no carry left, indicated by b becoming 0.

• Result: Once b is 0, it signifies that no further carry is being generated, and a contains the final computed sum. This sum is then returned as the solution.

Conclusion

This problem elegantly demonstrates how bit manipulation can effectively replicate basic arithmetic operations, such as addition, without using standard arithmetic operators. It offers a unique perspective on understanding and applying bitwise operators to solve computational problems, showcasing the versatility and power of bit-level operations in programming.

Rust Solution

Rust binds both parameters with mut a: i32 and mut b: i32 because bindings are immutable by default, and the loop reassigns a and b in place each iteration. The carry uses .wrapping_shl(1) rather than a plain shift: pushing a set bit into the sign bit of an i32 is arithmetic overflow, which panics in a debug build, so the wrapping variant makes the two's-complement wrap explicit and non-panicking. The body ends with a bare a and no return keyword — in Rust the trailing expression is the function's return value.

Go Solution

Go folds both parameters into a single a, b int group and reuses them as ordinary mutable locals, so no mut marker is needed to reassign a and b. for b != 0 is the entire loop — Go has no while, so a for with only a condition fills that role. carry := (a & b) << 1 declares the carry with the := short form, and the plain << silently wraps on int overflow instead of panicking, so unlike the Rust version no wrapping helper is required before the explicit return a.