LongAdder源码分析

一、LongAdder构造

1
2
3
4
5
6
    /**
     * Creates a new adder with initial sum of zero.
     */
    public LongAdder() {
    }

二、add

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
 /**
     * Adds the given value.
     *
     * @param x the value to add
     */
    public void add(long x) {
        Cell[] as; long b, v; int m; Cell a;
        if ((as = cells) != null || !casBase(b = base, b + x)) { 
            boolean uncontended = true;
            if (as == null || (m = as.length - 1) < 0 ||
                (a = as[getProbe() & m]) == null ||
                !(uncontended = a.cas(v = a.value, v + x)))
                longAccumulate(x, null, uncontended);
        }
    }

    /**
     * Equivalent to {@code add(1)}.
     */
    public void increment() {
        add(1L);
    }

    /**
     * Equivalent to {@code add(-1)}.
     */
    public void decrement() {
        add(-1L);
    }

  • 判断Cell数组是否为空
      • CAS更新base,这个一般情况就一个线程操作
        • 更新失败,调用复杂longAccumulate
    • 否,当前线程的探测值查找相应的Cell,Cell不存在或者CAS的value失败都调用longAccumulate

三、sum

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
    /**
     * 进行累加操作
     *
     * @return the sum
     */
    public long sum() {
        Cell[] as = cells; Cell a;
        long sum = base;
        if (as != null) {
            for (int i = 0; i < as.length; ++i) {
                if ((a = as[i]) != null)
                    sum += a.value;
            }
        }
        return sum;
    }

四、reset

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
   /**
     * 重置
     */
    public void reset() {
        Cell[] as = cells; Cell a;
        base = 0L;
        if (as != null) {
            for (int i = 0; i < as.length; ++i) {
                if ((a = as[i]) != null)
                    a.value = 0L;
            }
        }
    }

    /**
     *
     * @return the sum
     */
    public long sumThenReset() {
        Cell[] as = cells; Cell a;
        long sum = base;
        base = 0L;
        if (as != null) {
            for (int i = 0; i < as.length; ++i) {
                if ((a = as[i]) != null) {
                    sum += a.value;
                    a.value = 0L;
                }
            }
        }
        return sum;
    }