Java错误：构造函数未定义

在Java中，为什么会出现以下错误：

Error: The constructor WeightIn() is undefined

Java代码：

public class WeightIn{
  private double weight;
  private double height;

  public WeightIn (double weightIn, double heightIn){
    weight = weightIn;
    height = heightIn;
  }
  public void setWeight(double weightIn){
    weight = weightIn;
  }
  public void setHeight(double heightIn){
    height = heightIn;
  }
}

public class WeightInApp{
  public static void main (String [] args){
    WeightIn weight1 = new WeightIn();         //Error happens here.
    weight1.setWeight(3.65);
    weight2.setHeight(1.7);
  }
}

---

我定义了一个构造函数

将此添加到您的类中：

public WeightIn(){
}

public class WeightIn {
  ...

  public WeightIn (double weightIn, double heightIn){
    weight = weightIn;
    height = heightIn;
  }

  ...
}

• 请理解，只有在没有编写其他构造函数的情况下，才会提供默认的无参数构造函数
• 若您编写任何构造函数，那个么编译器不提供默认的无参数构造函数。您必须指定一个

---

您没有构造函数WeightIn（）。请创建它或将main方法中的参数提供给构造函数

WeightIn weight1 = new WeightIn();  

未定义默认构造函数。请这样定义它：-

public weightIn()
    {
    }

---

在这种情况下，您不能执行

WeightIn-weight1=new-WeightIn（）因为未定义默认构造函数

所以你可以加上

public WeightIn(){
}

或者你可以这样做

WeightIn weight1=new WeightIn（3.65,1.7）//构造函数接受两个双值
编译器在此行遇到对“
WeightIn（）
”无参数构造函数的调用：

WeightIn weight1 = new WeightIn();         //Error happens here.

编译器正在类定义中寻找匹配的构造函数，但没有找到它。这就是错误所在。（您确实定义了一个构造函数：“
WeightIn（double，double）
”，但它需要两个参数，并且不匹配。）

有几种方法可以解决这个问题

最简单的方法是更改main方法中的代码以传递两个参数

WeightIn weight1 = new WeightIn( 3.65, 1.7); 
//weight1.setWeight(3.65);
//weight2.setHeight(1.7);

对
setWeight
和
setHeight
方法的调用是多余的，因为构造函数方法已经为成员赋值。
首先，您应该知道一个.java文件只能有一个公共类

由于编写了参数化构造函数并访问了默认构造函数，所以出现了错误。要修复此错误，请写入：

WeightIn weight1 = new WeightIn(5.2, 52.2); 

而不是

WeightIn weight1 = new WeightIn();

这只是因为您的类没有匹配的构造函数：

public WeightIn(){
}

public class WeightIn {
  ...

  public WeightIn (double weightIn, double heightIn){
    weight = weightIn;
    height = heightIn;
  }

  ...
}

您需要添加
公共权重（）

在Java中，如果没有定义or，编译器会自动生成or。因此，当您定义以下类时：

public class WeightIn {
  private double weight;
  private double height;

  // didn't define a constructor.
  public void setWeight(double weightIn){
    weight = weightIn;
  }
  public void setHeight(double heightIn){
    height = heightIn;
  }
}

编译器将为您生成匹配的默认构造函数。因此，您的类隐式地具有如下默认构造函数：

public class WeightIn {
  private double weight;
  private double height;

  // automatically generated by compiler
  public WeightIn() {
    // do nothing here.
  }

  // didn't define a constructor.
  public void setWeight(double weightIn){
    weight = weightIn;
  }
  public void setHeight(double heightIn){
    height = heightIn;
  }
}

使用以下命令实例化类时：

WeightIn weight = new WeightIn(); 

public class WeightIn {
  ...

  // this won't automatically generated by compiler
  // public WeightIn() {
  //   nothing to do here.
  //}

  public WeightIn (double weightIn, double heightIn){
    weight = weightIn;
    height = heightIn;
  }

  ...
}

一切都很好

但是，当您自己添加构造函数时，编译器不会生成默认构造函数。因此，当您使用以下内容创建类时：

WeightIn weight = new WeightIn(); 

public class WeightIn {
  ...

  // this won't automatically generated by compiler
  // public WeightIn() {
  //   nothing to do here.
  //}

  public WeightIn (double weightIn, double heightIn){
    weight = weightIn;
    height = heightIn;
  }

  ...
}

您将没有默认构造函数（即
publicweightin（）{}
）。并且使用以下内容将引发错误，因为您没有匹配的构造函数：

 WeightIn weight = new WeightIn();

这花了我一段时间，但我想我终于想出了一个办法让这个程序工作。
我将这些类分为不同的文件，并将weight类重命名为Record，这样它就更明显了，而不是将所有内容都命名为weight的一些变体。我还添加了一个输出类，以尽可能保持主代码中的代码干净和最少。
我希望这和你希望做的一样。

原始代码：“构造函数”


项目：权重应用

套餐：weightInApp

类：Record.Java

package weightInApp;

        class Record 
        {
            private double weight; //declare variables
            private double height;
        Record (double weight, double height) {  //Parameterized Constructor method
                    this.weight = weight;  //this is the correct way 
                    this.height = height;
    //if you want to use weightIn and/or heightIn you have to be consistent acrosss 
    //the whole class. I decided to drop "In" off the variables for brevity.
                }
                //Getters and setters
                public double getWeight() { 
                    return weight;
                }
                public void setWeight(double weight) {
                    this.weight = weight;
                }
                public double getHeight() {
                    return height;
                }
                public void setHeight(double height) {
                    this.height = height;
                }
        }

package weightInApp;
import weightInApp.Record; //imports methods and variables from the other classes
import weightInApp.Output;
public class  Main {

    public static void main (String [] args){  
        Record weightIn1 = new Record(0,0);  
//previous line of code creates a new Record object named weightIn1 
//previous line of code also sets both the values of weight and height to 0.
        weightIn1.setWeight(3.65); //sets weight for weightIn1
        weightIn1.setHeight(1.70);//sets Height for WeightIn1
        Record weightIn2 = new Record(0, 0); 
        weightIn2.setWeight(3.00);
        weightIn2.setHeight(1.75);
 //previous 3 lines are the same as above for weightIn1 
//except this is for a second object named weightIn2
        Output.output(weightIn1, weightIn2); //calls runs passes values to output method
}
}


项目：权重应用

套餐：weightInApp

类：Output.Java

此类将设置的值输出到控制台上的表中。可以手动更改以添加更多数据。您还可以考虑跟踪记录的日期并将其添加到此输出中。您需要在记录类中为该功能添加必需的变量、getter和setter

package weightInApp;

public class Output {
    static void output (Record weightIn1, Record weightIn2)
    {
        int count = 0;
        final Object[][] table = new Object[3][3];
        table[0] = new Object[] { "Record", "Weight", "Height"};
        table[1] = new Object[] { count +=1, weightIn1.getWeight(), weightIn1.getHeight() };
        table[2] = new Object[] { count+=1, weightIn2.getWeight(), weightIn2.getHeight() };
        for (final Object[] row : table) {
            System.out.format("%15s%15s%15s\n", row);
        }

}
}


项目：权重应用

套餐：weightInApp

类：Main.Java

package weightInApp;

        class Record 
        {
            private double weight; //declare variables
            private double height;
        Record (double weight, double height) {  //Parameterized Constructor method
                    this.weight = weight;  //this is the correct way 
                    this.height = height;
    //if you want to use weightIn and/or heightIn you have to be consistent acrosss 
    //the whole class. I decided to drop "In" off the variables for brevity.
                }
                //Getters and setters
                public double getWeight() { 
                    return weight;
                }
                public void setWeight(double weight) {
                    this.weight = weight;
                }
                public double getHeight() {
                    return height;
                }
                public void setHeight(double height) {
                    this.height = height;
                }
        }

package weightInApp;
import weightInApp.Record; //imports methods and variables from the other classes
import weightInApp.Output;
public class  Main {

    public static void main (String [] args){  
        Record weightIn1 = new Record(0,0);  
//previous line of code creates a new Record object named weightIn1 
//previous line of code also sets both the values of weight and height to 0.
        weightIn1.setWeight(3.65); //sets weight for weightIn1
        weightIn1.setHeight(1.70);//sets Height for WeightIn1
        Record weightIn2 = new Record(0, 0); 
        weightIn2.setWeight(3.00);
        weightIn2.setHeight(1.75);
 //previous 3 lines are the same as above for weightIn1 
//except this is for a second object named weightIn2
        Output.output(weightIn1, weightIn2); //calls runs passes values to output method
}
}

样本输出

编译器是正确的。构造函数
WeightIn（）
未定义。请更正代码，然后通过一份副本，以便我能理解它。rit仍然不起作用+1这是唯一一个真正澄清错误原因的答案：它不在构造函数的定义中，它是在调用一个与任何已定义的构造函数签名都不匹配的构造函数。这种见解将问题的焦点从“我没有正确地编写类构造函数”转移到“我没有正确地调用类构造函数”，这对n00b（像我一样）对问题根源的理解产生了影响。所有其他答案似乎都暗示，在重载无参数构造函数之前，必须先编写无参数构造函数，但事实并非如此。@TomAuger:你完全正确。不需要提供无参数构造函数。我们可能想这样做，但这不是必须的。（你是对的，有些答案误导性地使用了“你需要”一词，并且只提供了一个解决方案，好像这是唯一的解决方案。（每当我读到一个回答说“你需要”我总是问，这是真的吗？一个人真的需要，或者有其他更好的选择吗？@TomAuger：你也很正确，这里的许多答案跳过了被问到的问题“为什么我会犯这个错误？”并尝试回答其他问题。