java - 简化便利构造函数

Question

我有一个工作方便的构造函数，但我觉得它的代码太多了。我不确定如何简化它，但我将不胜感激任何帮助!

public Hand(Card c1, Card c2, Card c3, Card c4, Card c5, Card c6) {
    this();
    this.card1 = c1;
    this.card2 = c2;
    this.card3 = c3;
    this.card4 = c4;
    this.card5 = c5;
    this.card6 = c6;
    if (c1 != null && c2 != null && c3 != null && c4 != null && c5 != null && c6 != null) {
        for (int count = 0; count < 6; count++) {
            if (count == 0) {
                cardsInHand.add(c1);
            } else if (count == 1) {
                cardsInHand.add(c2);
            } else if (count == 2) {
                cardsInHand.add(c3);
            } else if (count == 3) {
                cardsInHand.add(c4);
            } else if (count == 4) {
                cardsInHand.add(c5);
            } else if (count == 5) {
                cardsInHand.add(c6);
            }
        }
    }
}

编辑:根据下面的建议清理了代码。程序仍然运行，代码如下:

public Hand(Card c1, Card c2, Card c3, Card c4, Card c5, Card c6) {
    this();
    this.card1 = c1;
    this.card2 = c2;
    this.card3 = c3;
    this.card4 = c4;
    this.card5 = c5;
    this.card6 = c6;

    cardsInHand.add(c1);
    cardsInHand.add(c2);
    cardsInHand.add(c3);
    cardsInHand.add(c4);
    cardsInHand.add(c5);
    cardsInHand.add(c6);

Answer 1

for 循环是多余的，因为它将按该顺序添加

cardsInHand.add(c1);
cardsInHand.add(c2);
//etc

会做同样的事情。

您还可以考虑使用 var args 构造函数:

public Hand(Card ... cards){
   this.card1=cards[0];
   //etc