Java | 仅使用递归和条件创建显式加法函数

Question

前言

通过在日程安排中找到一些空闲时间，我要求自己提高自己的递归技能（不幸的是）。作为实践，我想通过使用递归来重新创建所有运算符，第一个是加法运算。虽然我有点卡住。

题

隐含地，我想仅通过使用递归和条件来重新创建加法运算符。尽管我完成了很大一部分代码，但由于只包含一个加法运算符，仍然存在一个问题。这是代码（可以正常运行，并且可以按预期在正，负和零输入的所有变体中添加）。我还提供了一些平庸的评论作为帮助。

public class Test {
    public static void main(String[] args) {
        // Numbers to add
        int firstNumb = -5, secondNumb = 3;
        // Call the add function and save the result
        int result = add(firstNumb, secondNumb);
        // Print result
        System.out.println(result);
    }

    /*
     * Function recursively takes a number from 'giver' one at a time and
     * "gives"/"adds" it to 'receiver'. Once nothing more to "give" (second == 0),
     * then return the number that received the value, 'receiver'.
     */
    public static int add(int receiver, int giver) {
        /*
         * Base Case since nothing more to add on. != to handle signed numbers
         * instead of using > or <
         */
        if (giver != 0) {
            /*
             * Recursive Call.
             * 
             * The new 'giver' param is the incremental value of the number
             * towards 0. Ex: -5 -> -4 , 5 -> 4 (so I guess it may decrement).
             * 
             * The new 'receiver' param is the incremental value based on the
             * opposite direction the 'giver' incremented (as to why the
             * directionalIncrement() function needs both values to determine
             * direction.
             */
            return add(directionalIncrement(receiver, giver),
                    directionalIncrement(giver, -giver));
        } else {
            // Return 'receiver' which now contains all values from 'giver'
            return receiver;
        }
    }

    // Increments (or decrements) the 'number' based on the sign of the 'direction'
    public static int directionalIncrement(int number, int direction) {
        // Get incremental value (1 or -1) by dividing 'direction' by absolute
        // value of 'direction'
        int incrementalValue = direction / abs(direction);
        // Increment (or decrement I guess)
        return number + incrementalValue;
    }

    // Calculates absolute value of a number
    public static int abs(int number) {
        // If number is positive, return number, else make it positive by multiplying by -1 then return
        number = (number > 0.0F) ? number : -number;
        return number;
    }
}

问题是包含的行return number + incrementalValue;。如前所述，尽管不符合我自己不涉及任何加法运算符的规范，但代码可与此一起使用。

我将该行更改为，return add(number, incrementalValue);但似乎无法打破递归规则，的确抛出了该网站的标题，StackOverflowException.

所有帮助表示赞赏。提前致谢。

注意

约束不包含任何隐式的增减（i ++ / i–），也不包含按位。尝试并回答我在自己的实现中遇到的特定问题。

Answer 1

public static int add(int a, int b) {
    if(b == 0) return a;
    int sum = a ^ b; //SUM of two integer is A XOR B
    int carry = (a & b) << 1;  //CARRY of two integer is A AND B
    return add(sum, carry);
}

从这里丢人。所有功劳归其作者。