Java BigDecimal、Integer、Long、Double类型数值累加求和

雨点打透心脏的1/2处 2024-03-27 09:49 78阅读 0赞

#### 文章目录 ####

*  1 BigDecimal类型数值累加求和
 *   *  1.1 for循环实现
     *  1.2 stream().reduce()实现
 *  2 Integer类型数值累加求和
 *   *  2.1 for循环实现
     *  2.2 stream().reduce()实现
     *  2.3 Collectors.summingInt()实现
     *  2.4 Collectors.summarizingInt()实现
 *  3 Long类型数值累加求和
 *   *  3.1 for循环实现
     *  3.2 stream().reduce()实现
     *  3.3 Collectors.summingLong()实现
     *  3.4 Collectors.summarizingInt()实现
 *  4 Double类型数值累加求和
 *   *  4.1 for循环实现
 *  5 [BigDecimal工具类][BigDecimal]
 *  6 [Object转BigDecimal类型][Object_BigDecimal]
 *  总结

## 1 BigDecimal类型数值累加求和 ##

### 1.1 for循环实现 ###

List<BigDecimal> list=new ArrayList<>();
    BigDecimal sum=new BigDecimal(0);
    for(BigDecimal decimal:list){
        
        sum=sum.add(decimal);
    }

### 1.2 stream().reduce()实现 ###

List<BigDecimal> list=new ArrayList<>();
    BigDecimal sum=list.stream().reduce(0,BigDecimal::add);

## 2 Integer类型数值累加求和 ##

### 2.1 for循环实现 ###

List<Integer> list=new ArrayList<>();
    Integer sum=0;
    for(Integer i:list){
        
        sum+=i;
    }

### 2.2 stream().reduce()实现 ###

List<Integer> list=new ArrayList<>();
    //lambda表达式实现
    Integer sum = list.stream().reduce(0, (current, number) -> current + number);
    //方法引用实现
    Integer sum2 = list.stream().reduce(0, Integer::sum);

### 2.3 Collectors.summingInt()实现 ###

List<Integer> list=new ArrayList<>();
    long sum = list.stream().collect(Collectors.summingInt(x -> x));

### 2.4 Collectors.summarizingInt()实现 ###

List<Integer> list = new ArrayList<>();
            IntSummaryStatistics summaryStatistics = list.stream().collect(Collectors.summarizingInt(x -> x));
            long sum = summaryStatistics.getSum();

## 3 Long类型数值累加求和 ##

### 3.1 for循环实现 ###

List<Long> list=new ArrayList<>();
    Long sum=0L;
    for(Long i:list){
        
        sum+=i;
    }

### 3.2 stream().reduce()实现 ###

List<Long> list=new ArrayList<>();
    //lambda表达式实现
    Long sum = list.stream().reduce(0L, (current, number) -> current + number);
    //方法引用实现
    Long sum2 = list.stream().reduce(0L, Long::sum);

### 3.3 Collectors.summingLong()实现 ###

List<Long> list=new ArrayList<>();
    long sum = list.stream().collect(Collectors.summingLong(x -> x));

### 3.4 Collectors.summarizingInt()实现 ###

List<Long> list = new ArrayList<>();
            LongSummaryStatistics summaryStatistics = list.stream().collect(Collectors.summarizingLong(x -> x));
            long sum = summaryStatistics.getSum();

## 4 Double类型数值累加求和 ##

### 4.1 for循环实现 ###

List<Double> list = new ArrayList<>();
    Double sum = 0d;
    for (Double i : list) {
        
        sum += i;
    }

> 注意 Double小数点失真问题解决 : **先把Double转为BigDecimal，再求和。代码如下 :**

List<Double> list = new ArrayList<>();
            list.add(1.24224D);
            list.add(1.24224D);
            list.add(1.32224D);
            BigDecimal bSum = new BigDecimal(0);
            for (Double i : list) {
        
                bSum = bSum.add(BigDecimal.valueOf(i));
            }
            System.out.println(bSum.doubleValue());

## 5 [BigDecimal工具类][BigDecimal] ##

## 6 [Object转BigDecimal类型][Object_BigDecimal] ##

--------------------

## 总结 ##

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[参考1][1]

[BigDecimal]: https://blog.csdn.net/weixin_42326851/article/details/129797911?csdn_share_tail=%7B%22type%22%3A%22blog%22%2C%22rType%22%3A%22article%22%2C%22rId%22%3A%22129797911%22%2C%22source%22%3A%22weixin_42326851%22%7D
[Object_BigDecimal]: https://blog.csdn.net/weixin_42326851/article/details/128780294?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522168040883716782425180885%2522%252C%2522scm%2522%253A%252220140713.130102334.pc%255Fblog.%2522%257D&request_id=168040883716782425180885&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~blog~first_rank_ecpm_v1~rank_v31_ecpm-3-128780294-null-null.blog_rank_default&utm_term=bigdecimal&spm=1018.2226.3001.4450
[1]: https://blog.csdn.net/qq_38974638/article/details/113913754