Question

问题是当我输入小数输入时。例如10001.10。它说NumberFormatException。但是，当我输入一个没有小数的数字，如＆＃34; 1001110＆＃34;它的工作正常。

import java.util.*;
import java.util.Arrays;


public class Binary {
public static void main (String[] args) 
{
    Scanner in = new Scanner(System.in);

    String[] Condition = { "0", "1", "." };

    //Accept First Input
    String numInp1="";      
    System.out.print("Enter First Binary Number: ");
    numInp1 = in.nextLine();

    //Accept Second Input
    String numInp2="";      
    System.out.print("Enter Second Binary Number: ");
    numInp2 = in.nextLine();

    //numInp1
    String _Array1 []=new String[numInp1.length()];
    //numInp2
    String _Array2 []=new String[numInp2.length()];

    //Catch error in numInp1/_Array1
     boolean flag1 = false;

     for(int i = 0; i < _Array1.length; i++)
     {
         _Array1[i] = String.valueOf(numInp1.charAt(i));
            int cnt = 0;
            for (int j = 0; j < Condition.length; j++)
            {
                if (!_Array1[i].equals(Condition[j]))
                {
                    cnt++;
                }
                else
                {
                    break;
                }
                if (cnt == 3)
                {
                    flag1 = true;
                }

            }

      }
    //Catch error in numInp2//_Array2
     boolean flag2 = false;

     for(int i = 0; i < _Array2.length; i++)
     {
           _Array2[i] = String.valueOf(numInp2.charAt(i));
           int cnt = 0;

            for (int j = 0; j < Condition.length; j++)
            {
                if (!_Array2[i].equals(Condition[j]))
                {
                    cnt++;
                }
                else
                {
                    break;
                }
                if (cnt == 3)
                {
                    flag2 = true;
                }
            }             
     }
     //Getting which of the Array has Higher Input
     if(flag2 == false && flag1 == false)
     {

        int HigherLength = 0;

        if(_Array1.length >= _Array2.length)
        {
            HigherLength = _Array1.length + _Array2.length;
        }
        else
        {
            HigherLength = _Array2.length + _Array1.length;
        }
    //Declaring the size of higher length as the size of Equals[] Array 
        int Equals[] = new int[HigherLength];   
        int _Array1Int[] = new int[_Array1.length];
        int _Array2Int[] = new int[_Array2.length];

      for(int i = 0;i<_Array1.length;i++)
      {
        if(_Array1[i].equals("."))
        {
            _Array1[i] = "6";

        }
        _Array1Int[i] = Integer.parseInt(_Array1[i]);
      }

      for(int i = 0;i<_Array2.length;i++)
      {
        if(_Array2[i].equals("."))
        {
            _Array2[i] = "6";

        }
        _Array2Int[i] = Integer.parseInt(_Array2[i]);
      }

     //_________________________

     int numInp1int = Integer.parseInt(numInp1);
     int numInp2int = Integer.parseInt(numInp2);




     int i,m,n,sum,carry=0;
    for(i=Equals.length-1;i>=0;i--)
    {
        m=numInp1int%10;
        n=numInp2int%10;
        numInp1int=numInp1int/10;
        numInp2int=numInp2int/10;
        sum=m+n+carry;  

        if(sum==1)
        {
          Equals[i]=1;
          carry=0;
        }

        else if(sum==2)
        {
            Equals[i]=0;
            carry=1;
        }

        else if(sum==3)
        {
            Equals[i]=1;
            carry=1;
        }

        else
        {
            Equals[i]=m+n+carry;
        }
    }

    String Equals1[] = new String[Equals.length];

    for(i=0;i<Equals.length;i++)
    {
        try{

        Equals1[i] = String.valueOf(Equals[i]);
        }
        catch (NumberFormatException nfe) 
        {
         nfe.printStackTrace();
        }
        if (Equals1[i].equals("6"))
        {
            Equals1[i] = ".";

        }
    System.out.print(Equals1[i]);
    }




    }
}
}

Answer 1

在代码中执行以下更改

删除以下2行 -

int numInp1int = Integer.parseInt(numInp1);
int numInp2int = Integer.parseInt(numInp2);

替换为以下代码 -

    StringBuilder strNum1 = new StringBuilder();
    for (int num : _Array1Int) 
    {
         strNum1.append(num);
    }
    int numInp1int = Integer.parseInt(strNum1.toString());


    StringBuilder strNum2 = new StringBuilder();
    for (int num : _Array2Int) 
    {
         strNum2.append(num);
    }
    int numInp2int = Integer.parseInt(strNum2.toString());

Answer 2

更新_Array1和_Array2不会更改您尝试解析为整数的numInp1int和numInp2int。

您应该学会解释堆栈跟踪中的错误消息。在这里，您可以清楚地看到问题所在：

Exception in thread "main" java.lang.NumberFormatException: For input string: "10001.1"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.parseInt(Integer.java:615)
    at q35640801.Binary.main(Binary.java:120)

它出现在班级parseInt()的电话Integer中。它无法解析像10001.10

这样的十进制数字

在JAVA中添加二进制数。 NumberFormatException的

2 个答案: