java - 安卓:Java代码输出 "NaN"

Question

所以这是我第一次开发的应用程序。当我出于某种原因在模拟器中运行此代码时，我得到一个“NaN”输出。该程序本质上是为了从多个选择(数量和价格的组合)中找到最低价格。我不知道我做错了什么。有什么建议吗？

(注意:仅当并非所有 EditText 字段中都有数字时才会输出 NaN)

主类:

import android.app.Activity;
import android.os.Bundle;
import android.widget.TextView;
import android.widget.EditText;
import android.widget.Button;
import android.view.View;
import android.view.View.OnClickListener;

public class worthit extends Activity implements OnClickListener{
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    Button b = (Button)this.findViewById(R.id.btn_calculate);
    b.setOnClickListener(this);
}

public void onClick(View v){

    //Declaring all of our variables that we will use in 
    //future calculations
    EditText price1 = (EditText)this.findViewById(R.id.price1);
    EditText price2 = (EditText)this.findViewById(R.id.price2);
    EditText price3 = (EditText)this.findViewById(R.id.price3);
    EditText price4 = (EditText)this.findViewById(R.id.price4);

    EditText quant1 = (EditText)this.findViewById(R.id.quant1);
    EditText quant2 = (EditText)this.findViewById(R.id.quant2);
    EditText quant3 = (EditText)this.findViewById(R.id.quant3);
    EditText quant4 = (EditText)this.findViewById(R.id.quant4);

    //TextView box used to present the results
    TextView tv = (TextView)this.findViewById(R.id.result);

    //Declaring two arrays of the values from
    //all of our EditText fields
    double[] price = new double[4];
    double[] quantity = new double[4];

    try{
        price[0] = Double.parseDouble(price1.getText().toString());
        price[1] = Double.parseDouble(price2.getText().toString());
        price[2] = Double.parseDouble(price3.getText().toString());
        price[3] = Double.parseDouble(price4.getText().toString());

        quantity[0] = Double.parseDouble(quant1.getText().toString());
        quantity[1] = Double.parseDouble(quant2.getText().toString());
        quantity[2] = Double.parseDouble(quant3.getText().toString());
        quantity[3] = Double.parseDouble(quant4.getText().toString());

        if
    } catch(NumberFormatException nfe) {
        tv.setText("Parsing Error");
    }

    //Creating a Optimize class and using our
    //price and quantity arrays as our parameters
    Calculate optimize = new Calculate(price, quantity);

    //Calling the optimize method to compute the cheapest 
    //choice 
    optimize.optimize();

    //Composing a string to display the results
    String result = "The best choice is the $" +
            optimize.getResultInDollars() + " choice.";

    //Setting the TextView to our result string
    tv.setText(result);


   }
 }

这是我的类(class)，负责所有的处理工作:

//Work class used for computing whether
//one choice is cheaper than another given
//a choice of several options af different 
//prices and quantities
//Ex. Coffee- $1-10oz, $1.2-12oz, $1.4-16oz

public class Calculate {
//declaring variables
private double[] dollarValue;
private double[] ounce;
private int indexNumber;        //Index number of the lowest ratio
private double minValue;        //Lowest ratio

private double resultInDollars;
private double resultInOunces;


//class constructor
public Calculate(double[] dol, double[] oun){
    //initializing our variables
    dollarValue=new double[dol.length];
    ounce=new double[oun.length];

    //passing the values from the parameter
    //arrays in our arrays
    for(int i=0;i < dol.length;i++){
        dollarValue[i]=dol[i];
        ounce[i]=oun[i];
    }
}

//Optimize method used to compute the
//cheapest price per quantity
public void optimize(){
    //finding the ratio of the dollar value
    //and the quantity (ounces)
    double[] ratio=new double[dollarValue.length];
    for(int i=0;i<dollarValue.length;i++)
        ratio[i]=dollarValue[i]/ounce[i];

    //finding the smallest value in the ratio
    //array and its location (indexNumber)
    minValue = ratio[0];
    for(int i=1;i < dollarValue.length; i++){
        if(ratio[i] < minValue){
            minValue=ratio[i];
            indexNumber=i;
        }
    }

    //finding the dollar value of the smallest
    //ratio that we found above
    //e.g. most cost effective choice 
    setResultInDollars(minValue*ounce[indexNumber]);
    setResultInOunces(ounce[indexNumber]);
}

public void setResultInDollars(double dollarValueChoiche) {
    this.resultInDollars = dollarValueChoiche;
}

public void setResultInOunces(double resultInOunces) {
    this.resultInOunces = resultInOunces;
}

public double getResultInDollars() {
    return resultInDollars;
}

public double getResultInOunces() {
    return resultInOunces;
}

}

干杯。

编辑:显然我似乎在某处也有逻辑错误。例如，如果我选择以下价格:1.4、1.6，并且我选择以下数量(分别)18、20，输出告诉我 1.6 美元是最佳选择；当您手动计算 (1.4/18, 1/6,20) 时，您会发现 1.4 的比率最低，因此它必须是最佳选择。如果有人能告诉我我做错了什么，将不胜感激。

谢谢。

Answer 1

NaN 代表Not a Number并且是一个浮点占位符值，在您尝试计算无效计算时使用。

最常见的此类操作是除以零。因为我在你的代码中看到了一个分区 (ratio[i]=dollarValue[i]/ounce[i];)，所以我猜 ounce[i]在这一点上是 0。

请注意，您不能使用 == 检查 NaN，因为 NaN 不等于 任何值(甚至不是它本身!)。

要检查 float/double 是否为 NaN 使用 Float.isNaN()和 Double.isNaN()分别。