java - 有两个结构相同但细节不同的函数。如何摆脱重复？

Question

所以，这是我的功能:

  private void sendLeft() {
    leftSendersIndexes = newLeftSendersIndexes;
    Agent rightRecepient;
    int rightRecepientIdx = 0;
    Agent leftSender;
    for (int i = 0; i < leftSendersIndexes.size(); i++) {
        rightRecepientIdx = leftSendersIndexes.get(i) + 1;
        rightRecepient = list.get(rightRecepientIdx);
        leftSender = list.get(rightRecepientIdx - 1);
        rightRecepient.setNewLeftMsg(leftSender.getLeftMsg());
        rightRecepient.setLeftMsg(0); // reset left messages
    }
}

private void sendRight() {
    rightSendersIndexes = newRightSendersIndexes;
    Agent leftRecepient;
    int leftRecepientIdx = 0;
    Agent rightSender;
    for (int i = 0; i < rightSendersIndexes.size(); i++) {
        leftRecepientIdx = rightSendersIndexes.get(i) - 1;
        leftRecepient = list.get(leftRecepientIdx);
        rightSender = list.get(leftRecepientIdx + 1);
        leftRecepient.setNewRightMsg(rightSender.getRightMsg());
    }
}

它们非常相似。问题是，在第一个函数中，我有 leftRecepientIdx+1，然后是 leftRecepientIdx-1，我有 leftRecepientIdx-1 和 leftRecepientIdx +1 在第二个函数中。我可以将两个函数合二为一并添加一个 boolean 参数。但是有没有更好的方法来消除重复？

Answer 1

一种方法是使用这种重构:

private void sendLeft() {
    leftSendersIndexes = newLeftSendersIndexes;
    send(leftSendersIndexes, -1);
}

private void sendRight() {
    rightSendersIndexes = newRightSendersIndexes;
    send(rightSendersIndexes, +1);
}

private void send(List<Integer> indexes, int direction) {
    for (int i = 0; i < indexes.size(); i++) {
        int recipientIdx = indexes.get(i) - direction;
        Agent recipient = list.get(recipientIdx);
        Agent sender = list.get(recipientIdx + direction);
        if (direction == -1) {
            recipient.setNewLeftMsg(sender.getLeftMsg());
            recipient.setLeftMsg(0); // reset left messages
        }
        else {
            recipient.setNewRightMsg(sender.getRightMsg());
        }
    }
}

send 方法封装了基于direction 参数的逻辑:+1 为右，-1 为左。