java - 为什么即使该调用位于if语句内部，该递归方法仍会继续调用自身？

我正在研究一个我设计一种递归地计算字符串中x出现的方法的问题。从代码下面的输出中可以看到，当i == 6时，也包含递归调用的if条件不再执行。但是，为什么要多次打印“ countKeep is 4”？
此外，当我取消注释时，行数= 0;我只收到一次“ countKeep为4”，而以下几行打印“ countKeep为0”并返回0。

据我了解，变量countKeep应该设置为变量count。然后，将变量count和i设置为以下字符串的0。然后返回countKeep并给出该字符串中x的出现。

我对这个初学者的问题缺少什么？

我通过以下方式调用该方法：

    System.out.println(p.countX("xxhixx"));

    int count;
    int countKeep;
    int i;

    public int countX(String str) {
      if (i < str.length()) {
          System.out.println("i in outer if is " +i);
            if (str.charAt(i) == 'x') {
              count++;
              System.out.println("count in inner if is " + count);
            }
          i++;
          System.out.println("i is " +i);
          countX(str);
      }
      countKeep = count;
      System.out.println("countKeep is " + countKeep);
//    count = 0;
//    i = 0;
      return countKeep;
    }

输出：

i in outer if is 0
count in inner if is 1
i is 1
i in outer if is 1
count in inner if is 2
i is 2
i in outer if is 2
i is 3
i in outer if is 3
i is 4
i in outer if is 4
count in inner if is 3
i is 5
i in outer if is 5
count in inner if is 4
i is 6
countKeep is 4
countKeep is 4
countKeep is 4
countKeep is 4
countKeep is 4
countKeep is 4
countKeep is 4
4

最佳答案

在达到countKeep的设置并进行打印之前，请再次调用countX（）方法。因此，只有在完成对countX（）的所有调用后，它才会到达此代码（if语句之后的代码）。因此，它也被称为6次。

以下代码将更好地显示实际情况：

System.out.println(p.countX("xxhixx"));

int count;
int countKeep;
int i;

public int countX(String str) {
  if (i < str.length()) {
      System.out.println("i in outer if is " +i);
        if (str.charAt(i) == 'x') {
          count++;
          System.out.println("count in inner if is " + count);
        }
      i++;
      System.out.println("i is " +i);
      countKeep = count;
      System.out.println("countKeep is " + countKeep);
      countX(str);
  }
  System.out.println("Finished the countX() method");
  return countKeep;
}