我该如何解决这个问题我想将十进制数转换为二进制数

Question

public class Binar{
public static void main(String[] args){
    int num = 7;
    long Binary = cBtD(num);
System.out.printf("%d numri decimal = %d binar" , num, Binary);
}
public static long cBtD(int num){
    long BinaryNumber = 0;
    int  i = 0;
    long reminder;
    while(num > 0){
        reminder = num % 2;
        num /= 2;
        ++i;
    }
  for (int j = i - 1; j >= 0; j--) {
        System.out.print(BinaryNumber[j]); 
  }
    return BinaryNumber;
}}

and i have this error and it says "array required, but long found" and "System.out.print(BinaryNumber[j]);"

Answer 1

Reason behind this error is, you have defined BinaryNumber variable as long and it is not an array. But you are trying to access it like an array. Please check my modified answer below:

public class Binar {
public static void main(String[] args) {
    int num = 7;
    String Binary = cBtD(num);
    System.out.printf("%d numri decimal = %s binar", num, Binary);
}

public static String cBtD(int num) {
    String BinaryNumber = "";
    long reminder;
    if (num == 0) {
        return "0";
    }
    while (num > 0) {
        reminder = num % 2;
        BinaryNumber = String.valueOf(reminder).concat(BinaryNumber);
        num /= 2;
    }
    return BinaryNumber;
}
}

Answer 2

That error occurred because you defined BinaryNumber's type 'long' and you wanted use it as an array.

I change it a bit, try it:

public class Binar {

    public static void main(String[] args) {
        int num = 7;
        int[] binaryArray = cBtD(num);
        String numbers = "";
        for (int aBinaryArray : binaryArray)
            numbers += aBinaryArray;

        System.out.printf("%d numri decimal = %d binar" , num, Integer.parseInt(numbers));
    }

    private static int[] cBtD(int num){
        int i = 0;
        int temp[] = new int[7];
        int binaryNumber[];
        while (num > 0) {
            temp[i++] = num % 2;
            num /= 2;
        }
        binaryNumber = new int[i];
        int k = 0;
        for (int j = i - 1; j >= 0; j--) {
            binaryNumber[k++] = temp[j];
        }
        return binaryNumber;
    }
}

Or you can simply use these methods to convert decimal to binary:

Integer.toBinaryString();

Or this:

Integer.toString(n,2);

Answer 3

All numbers are inherently binary. But whether you display them in binary or hex or octal is simply a matter of representation. Which means you want to print them as a string. Even when you do the following:

int v = 123;
System.out.println(v); // v is printed as a decimal string.

So to convert them to a binary string, just prepend the remainders to the string after dividing by two (via the remainder operator).

int n = 11;
String s = "";

s = (n%2) + s; // s = "1"
n/=2;  // n == 5
s = (n%2) + s; // s = "11"
n/=2   // n == 2
s = (n%2) + s; // s = "011";
n/=2   // n == 1
s = (n%2) + s; // s = "1011";
n/=2;  // n == 0

n == 0 so your done.

return s and print it.