Question

该程序应该将二进制数转换为十进制数，并在输入具有非二进制数时抛出异常。这个程序将读取1，但是当我输入0时，它将抛出异常并告诉我它不是二进制文件。

测试程序：

//Prepare scanner from utility for input.
import java.util.Scanner;

public class Bin2Dec {
    public static void main (String[] args){
        //Convert the input string to their decimal equivalent.
        //Open scanner for input.
        Scanner input = new Scanner(System.in);
        //Declare variable s.
        String s;

        //Prompt user to enter binary string of 0s and 1s.
        System.out.print("Enter a binary string of 0s and 1s: ");
        //Save input to s variable.
        s = input.nextLine();
        //With the input, use try-catch blocks.
        //Print statement if input is valid with the conversion.
        try {
            System.out.println("The decimal value of the binary number "+ "'" + s + "'" +" is "+conversion(s));
            //Catch the exception if input is invalid.
        } catch (BinaryFormatException e) {
            //If invalid, print the error message from BinaryFormatException.
            System.out.println(e.getMessage());
        }
    }
    //Declare exception.
    public static int conversion(String parameter) throws BinaryFormatException {
        int digit = 0;
        for (int i = parameter.length(); i > 0; i--) {
            char wrong_number = parameter.charAt(i - 1);
            if (wrong_number == '1') digit += Math.pow(2, parameter.length() - i);
            //Make an else statement and throw an exception.

            else if (wrong_number == '0') digit += Math.pow(2, parameter.length() - i);

            else 
                throw new BinaryFormatException("");
        }
        return digit;
    } 
}

Answer 1

由于这些行，该程序仅接受'1'作为char：

if (wrong_number == '1') digit += Math.pow(2, parameter.length() - i);
          //Make an else statement and throw an exception.
else 
    throw new BinaryFormatException("");

由于没有if(wrong_number == '0')，该数字只接受1，并在遇到0时抛出异常。

除此之外：如果可能的话，尽量避免使用Math.pow。因为它是资源密集型的，在这种情况下完全没用。使用位移可以更容易地生成2 ^ x：

int pow_2_x = (1 << x);

最后：java已为此提供method：

int dec = Integer.parseInt(input_string , 2);

Answer 2

问题在于你的逻辑。由于您正在处理二进制数字（＆＃39; 1＆＃39;和＆＃39; 0＆＃39;）但您只检查1，您应该检查＆＃39; 0＆＃39; 0＆＃39; 0并且只有在＆＃39; 0＆＃39;以外的情况下才会抛出异常。 ans＆＃39; 1＆＃39;

if (wrong_number == '1') digit += Math.pow(2, parameter.length() - i);
//Make an else statement and throw an exception.


else 
    throw new BinaryFormatException("");

如何使我的二进制到十进制转换程序也读取0？

2 个答案: