Long 相等对比分析

四种对比方案

小于128L可以直接用==判断，存在风险不建议直接使用==

    Long a = 200L;
        Long b = 200L;
        System.out.println(a.equals(b));
        System.out.println(a.longValue()==b.longValue());
        System.out.println(a-b==0);
        System.out.println(a.compareTo(b) == 0);

检测项目代码时发现了另外的一种写法

oneBigNum.toString().equals(twoBigNum.toString()

一时兴起，对性能进行了一下测试（有点孔乙己的感觉）

package org.example;

public class Main {
    public static void main(String[] args) {
        Long oneNum=100L;
        Long twoNum=100L;
        Long oneCommNum127=127L;
        Long twoCommNum127=127L;

        Long oneCommNum128=128L;
        Long twoCommNum128=128L;
        Long oneBigNum=10000L;
        Long twoBigNum=10000L;

        int times=100*100*10000;

        System.out.println("equals times---> "+times);

        long t1=System.currentTimeMillis();
        for (int i=0;i<times;i++){
            boolean a=oneBigNum.equals(twoBigNum);
        }
        long t2=System.currentTimeMillis();

        System.out.println("oneBigNum.equals(twoBigNum)--耗时---> "+(t2-t1));
        t1=System.currentTimeMillis();
        for (int i=0;i<times;i++){
            boolean a=oneBigNum.toString().equals(twoBigNum.toString());
        }
        t2=System.currentTimeMillis();

        System.out.println("oneBigNum.toString().equals(twoBigNum.toString())--耗时---> "+(t2-t1));

        t1=System.currentTimeMillis();
        for (int i=0;i<times;i++){
            boolean a=oneBigNum.longValue()==twoBigNum.longValue();
        }
        t2=System.currentTimeMillis();

        System.out.println("oneBigNum.longValue()==twoBigNum.longValue()--耗时---> "+(t2-t1));

        t1=System.currentTimeMillis();
        for (int i=0;i<times;i++){
            boolean a=0==oneBigNum-twoBigNum;
        }
        t2=System.currentTimeMillis();

        System.out.println("0==oneBigNum-twoBigNum--耗时---> "+(t2-t1));

        t1=System.currentTimeMillis();
        for (int i=0;i<times;i++){
            boolean a=0==oneBigNum.compareTo(twoBigNum);
        }
        t2=System.currentTimeMillis();

        System.out.println("oneBigNum.compareTo(twoBigNum)--耗时---> "+(t2-t1));
    }
}

输出

equals times---> 100000000
oneBigNum.equals(twoBigNum)--耗时---> 6
oneBigNum.toString().equals(twoBigNum.toString())--耗时---> 6005
oneBigNum.longValue()==twoBigNum.longValue()--耗时---> 4
0==oneBigNum-twoBigNum--耗时---> 4
oneBigNum.compareTo(twoBigNum)--耗时---> 4