相同的构造函数名称错误

Question

好的，所以我正在尝试创建多个具有相同名称的构造函数。我收到一个错误，我不知道为什么。我基本上是在尝试这样做，以便当玩家输入他们的名字时，即使他们没有输入特定字段（例如后缀等），它仍然可以工作.我知道我仍然需要稍后添加setter和getter，但我只是想立即设置构造函数。关于为什么它给我一个错误的任何想法？

public class Name {
    private String firstName;
    private String middleInitial;
    private String lastName;
    private String suffix;


    public String playerName(String firstName, String middleInitial, String lastName, String suffix){

        this.firstName = firstName; 
        this.middleInitial = middleInitial;
        this.lastName = lastName;
        this.suffix = suffix;
        return suffix + firstName + middleInitial + lastName;
    }

    public String playerName(String firstName, String lastName){

        this.firstName = firstName; 
        this.middleInitial = middleInitial;
        this.lastName = lastName;
        this.suffix = suffix;
        return suffix + firstName + middleInitial + lastName;
    }

    public String playerName(String firstName, String middleInitial, String lastName){ //this is where the error is. "Duplicate method playerName (String String String) in type Name"

        this.firstName = firstName; 
        this.middleInitial = middleInitial;
        this.lastName = lastName;
        this.suffix = suffix;
        return suffix + firstName + middleInitial + lastName;
    }

    public String playerName(String firstName, String lastName, String suffix){//same error here

        this.firstName = firstName; 
        this.middleInitial = middleInitial;
        this.lastName = lastName;
        this.suffix = suffix;
        return suffix + firstName + middleInitial + lastName;
    }

Answer 1

当重载方法（或构造函数）时，您只能使用类型签名不同的方法。因此，只有一种方法（同名）采用三个String 参数。但是，您目前没有构造函数。我认为您真正想要的是创建Name 实例，也许还有getter 和setter，最后覆盖toString（这样您就可以获得Name 作为String）。类似的，

public class Name {
    private String firstName;
    private String middleInitial = "";
    private String lastName;
    private String suffix = "";

    public Name(String firstName, String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public Name(String firstName, String middleInitial, String lastName) {
        this(firstName, lastName);
        this.middleInitial = middleInitial;
    }

    public Name(String firstName, String middleInitial, String lastName, 
                String suffix) {
        this(firstName, lastName, middleInitial);
        this.suffix = suffix;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getMiddleInitial() {
        return middleInitial;
    }

    public void setMiddleInitial(String middleInitial) {
        this.middleInitial = middleInitial;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getSuffix() {
        return suffix;
    }

    public void setSuffix(String suffix) {
        this.suffix = suffix;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append(firstName);
        if (sb.length() > 0) {
            sb.append(' ');
        }
        if (middleInitial.length() > 0) {
            sb.append(middleInitial).append(' ');
        }
        sb.append(lastName);
        if (suffix.length() > 0) {
            sb.append(' ');
            sb.append(suffix);
        }
        return sb.toString();
    }
}

示例toString 正在执行String 连接，您可以做的效率较低

@Override
public String toString() {
    String str = firstName;
    if (str.length() > 0) {
        str += " "; // <-- str = str + " ";
    }
    if (middleInitial.length() > 0) {
        str += middleInitial + " ";
    }
    str += lastName;
    if (suffix.length() > 0) {
        str += " " + suffix;
    }
    return str;
}

Answer 2

让我们在你有错误的地方使用构造函数

public String playerName(String firstName, String middleInitial, String lastName) { /* error */ }

现在这是另一个你有同样错误的地方

public String playerName(String firstName, String lastName, String suffix){ /* same error */ }

那是因为您对它们中的参数和相同数量的参数使用相同的 DATA TYPE。因此，当您使用其中之一时，java 将无法决定调用哪一个。 我希望你现在已经清楚了......

Answer 3

如果你打电话

obj.playerName ("string1", "string2", "string3");

java怎么知道这是否是

public String playerName(String firstName, String middleInitial, String lastName)

或

public String playerName(String firstName, String lastName, String suffix)