这是我就该主题提出的最后一个问题的后续措施。但是,这是一个不同的问题。
我的代码正常工作,除了它使用copyOfRange复制某种地址。由于某种地址,它总是返回0.0,而不是数组getBits的部分。
有人可以扫描一下并提出建议吗?我为此感到疯狂(这不是一项任务)。
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);
}
}
}
}
最佳答案
将convert = tempBinary.toString();
行替换为:
convert = "";
for(int bin : tempBinary){
convert += bin;
}
那应该使您的转换正常工作。