從解析字符串到雙精度的轉換返回0.0：

Question

這是我就該主題提出的最后一個問題的后續措施。 但是，這是一個不同的問題。

我的代碼正常工作，除了它使用copyOfRange復制某種地址。 由於某種地址，它總是返回0.0，而不是數組getBits的部分。

有人可以掃描一下並提出建議嗎？ 我為此感到瘋狂（這不是一項任務）。

package runTests;

import java.util.Arrays;

public class runTestGetBinaryStrands {
    protected static int getBits[] = {1,0,1,1,0,1,0,0,0,1,1,0,1,0,1,0};
    double numerator, denominator, x, y;

    public static void main (String[] args){
        runTestGetBinaryStrands test = new runTestGetBinaryStrands();
        test.getNumber(null, getBits);
    }
    /*NOTE OF THIS FORLOOP:     * Divided the bits array in half & convert two different binary values to a string  * I parsed the string to an int value, which can be put saved to a double and be treated like a decimal value.  * I got the first 8 elements and stashed them into numerator, and did the same for denominator for the remaining array bits.    * * The chromosome has one binary string, made up of a bunch of smaller parts.* You use getNumber in the chromosome to get out the values of the parts.     **/     

    public void getNumber(String convert, int[] tempBinary){       
        for (int i = 0; i < getBits.length; i++){
            for(int j = 0; j < getBits.length; j++){ //start at index 0 to 7 = 8.
                tempBinary = Arrays.copyOfRange(getBits, 0, 7); //Get first set of 8 elements.
                convert = tempBinary.toString();
                System.out.println(convert);
                try{
                    numerator = Integer.parseInt(convert); //converts string to one whole section in
                }catch (NumberFormatException ex){
                }
                System.out.println("See Numerator's value: " + numerator);                          
                tempBinary= Arrays.copyOfRange(getBits, 8, 15); //Get Second set of 8 elements.
                convert = tempBinary.toString();
                try{
                    denominator = Integer.parseInt(convert); //converts string to one whole section in
                }
                catch (NumberFormatException ex){
                }                           
                System.out.println("See Denominator's value: " + denominator);
            } 
        } 
    } 
}

Answer 1

替換行convert = tempBinary.toString(); 有：

    convert = "";
    for(int bin : tempBinary){
      convert += bin;
    }

那應該使您的轉換正常工作。