两个短整数的异或 [英] XOR of two short integers

问题描述

我正在使用 XOR ^ 短整数的 XOR $ c>以传统方式运营。以下是方法 -

I am calculating XOR of two short integers using XOR ^ operator in a traditional fashion. Below is the method-

short a=197;
short b=341;
short y = (short) (a ^ b);

然而，XOR总是返回整数但在我的情况下输入是短整数，这就是为什么我要投XOR输出短路。 XOR可以用不同的方式计算（例如：使用 BigInteger 等），但性能明智（更少的时间），这对短整数最好？在记住性能的同时，我应该首先使用 Integer.toBinaryString（number）将每个短整数转换为二进制数，然后应用按位XOR？

However the XOR always returned integer but in my case inputs are short integer, that is why i am casting short to the XOR output. The XOR can be calculated in different manners (example: using BigInteger etc.) but performance wise (less time) which is the best for short integers? While keeping performance in mind, should i first convert each short integer to binary number using Integer.toBinaryString(number) then apply bitwise XOR?

推荐答案

short s1 = ...
short s2 = ...
short result = (short) (s1 ^ s2);

这是XOR两个短在一起。它不会遇到创建 BigInteger 的开销，并且转换将永远不会导致溢出问题，因为 s1 和 s2 以短开头。

This is the most efficient way to XOR two shorts together. It does not run into the overhead of creating BigIntegers and the cast will never cause an overflow issue as both s1 and s2 are shorts to begin with.

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