从 ArrayList <Integer>构建十六进制字符串

Question

我已经计算了余数，但我的问题是从余数到新的十六进制字符的实际转换。在 Dec2Hex 类中，我需要做的就是从超类中获取余数组列表(我已经这样做了(，然后使用字符数组 hexDigits[] 将其转换为字符串。我必须在循环中逐步遍历其余数组列表并构建十六进制字符串，但我不确定如何做到这一点。

//this is the Binary to Decimal implementation which hex2dec is based off of
package business;
import java.util.ArrayList;
/**
 *
 * @author
 */
public class Bin2Dec implements Conversion{
    public static final String VALUEDESC ="Binary";
    public static final String RESULTDESC ="Decimal";
    private String origval, result;
    private ArrayList<String> resultsteps;
    private String emsg;
    private boolean valid;
    public Bin2Dec(String value) {
        emsg = "";
        origval = value;
        if (valid =isValid(value)) {  //valid will be true or false
            convert(value);
        } else {
            emsg = "Illegal, binary value (must be only zeros and ones";
        }
    }
    private boolean isValid(String v) {
        for (int i=0; i< v.length(); i++) {
            if (v.charAt(i) != '1' && v.charAt(i) != '0') {
                return false;
            }
        }
        return true;
    }
    // we can do isValid because we declared private boolean valid
    @Override
    public boolean isValid() {
        return this.valid;
    }
    private void convert(String v) {
        long r = 0;
        String reverse = new StringBuilder(v).reverse().toString();
        resultsteps = new ArrayList<>();
        for (int i=0; i < reverse.length(); i++) {
            if (reverse.charAt(i) == '1') {
                long p = (long) Math.pow(2,i);
                r += p;
                resultsteps.add("There is a(n) " + p + " in the number (2^" + i + ")" );
            }
        }
        this.result = String.valueOf(r);
    }
    @Override
    public String getResult() {
        return this.result;
    }
    @Override
    public ArrayList<String> getProcessLog() {
        return this.resultsteps;
    }
    @Override
    public String getErrorMsg() {
        return this.emsg;
    }
    @Override
    public String getValue() {
        return this.origval;
    }
}

实现

//this is the hex2dec implementation I created 
package business;
import java.util.ArrayList;
/**
 *
 * @author 
 */
public class Dec2Hex extends Dec2Num {
    public static final String VALUEDESC="Decimal";
    public static final String RESULTDESC="Hexadecimal";
//    private ArrayList<String> codes;
    public static final int BASE=16;
    private String binaryresult;
    char[] hexDigits = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
              'A', 'B', 'C', 'D', 'E', 'F'};
    String hex = "";
    public Dec2Hex(String value) {
        super(value, Dec2Hex.BASE);
        this.binaryresult="";  
    }

    @Override
    public String getResult() {
        if(!super.isValid()) {return "Invalid Value";}
        for(Integer i : super.getRemainders()) {                             //we deleted remainders in Dec2Bin so we call super
            String hex = Integer.toHexString(i);
            this.binaryresult += String.valueOf(i);
        }
        switch (hex) {

        }
       return this.binaryresult;
    }
//    public ArrayList<String> getHexConversion(int d) {
//        char[] hexDigits = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
//              'A', 'B', 'C', 'D', 'E', 'F'};
//        String hex = "";
//        while (d > 0) {
//            int digit = d % BASE;              
//            hex = hexDigits.charAt(digit) + hex; 
//            d = d / BASE;
//        }
//    return codes;
//    }
    //    public ArrayList<String> getHexConversion(int d) {
//
//        String hex = "";
//        while (d > 0) {
//            int digit = d % BASE;             
//            char hexDigits = (hex <= 9 && hex >= 0);
//            (char)(hexDigits + '0') : (char)(hexDigits- 10 + 'A');
//        }
//    return this.binaryresult;
//    }
}

接口和抽象实现

//interface
public interface Conversion {
    public String getValue();
    public String getResult();
    public boolean isValid();
    public ArrayList<String> getProcessLog();
    public String getErrorMsg();
    //each subclass should fulfill these methods   
}
//abstract implementation
public abstract class Dec2Num implements Conversion{
    private String origval, emsg;
    private boolean valid;
    private int base;
    private ArrayList<String> resultsteps;
    private ArrayList<Integer> remainders;
    public Dec2Num(String value, int base) {  //int base b/c it needs to know what to divide down by
       emsg ="";
       origval = value;
       this.base = base;
       try {
           long n = Long.parseLong(value);
           if (n < 0 ) {
               emsg = "Bad decimal value: must be positive.";
               this.valid = false;
           } else {
              this.valid = true;
              resultsteps = new ArrayList<>();
              remainders = new ArrayList<>();
              convertByRecur(n);
           }
        } catch (NumberFormatException e) {
           emsg = "Illegal value: not a decimal integer";
           this.valid = false;
        }
    }//end of constructor
        @Override
    public boolean isValid() {
        return this.valid;
    }
    private void convertByRecur(long n) {
        int r = 0;
        r = (int)(n % this.base);
        long newval = n / this.base;
        resultsteps.add(n + " divided by " + this.base + "="
                        + newval + " w/remainder of: "  +r);
        if (newval > 0) {
            convertByRecur(newval);  //recursive call
        } 
        remainders.add(r);
    }
    @Override
    public ArrayList<String> getProcessLog() {
        return this.resultsteps;
    }
    @Override
    public String getErrorMsg() {
        return this.emsg;
    }
    @Override
    public String getValue() {
        return this.origval;
    }
    @Override
    public abstract String getResult();
    protected ArrayList<Integer> getRemainders() {
        return this.remainders;
    }
}

Answer 1

首先，调用Integer.toHexString()应该有效，但它绕过了hexDigits数组的想法。唯一的潜在问题是它会给你小写字母，a尽管f你可能想要大写的等效字母。

String.valueOf(i)不正确。如果i是 7，它会给你7，但如果i 11，它会给你11，你想要B，十六进制数字。你可以只做

        this.binaryresult += hex;

你根本不需要你的switch声明。

假设您确实想使用hexDigits数组：

• 要从数组中获取正确的char，只需使用 hexDigits[i] 。

• 您可以将char直接追加到String，而无需先将其转换为String：

      this.binaryresult += yourChar;
  

在提供我的代码版本之前，我会让你稍微挣扎一下。我相信你能做到。