Trouble converting decimal value to binary

I'm making a program that converts decimal values to binary and vice versa. I've figured out how to convert binary to decimal, however I am having trouble with decimal to binary. I've seen many sources with code on how to do it but most of them involve using an array. I can't use an array because this program has a GUI and JLabel can't print the array. I've found a code that works with certain numbers, but doesn't with others. Here's the code:

public void convertBinary (int decimal)
    {
        String binary = "";
        int remainder;
        
        while (decimal != 0)
        {
            remainder = decimal % 2;
            decimal /= 2;
            binary += remainder;
        }
        lblDecBinAns.setText(String.valueOf(binary));
                
        
    }

The given number is inputted by the user ("decimal") which is taken when they press the button. I don't know if there's an adjustment that can be made to this code for it to work properly. Perhaps there's an entirely different algorithm that would work for this or maybe a way to print an array with JLabel. I've been stumped on this for a while so any help is appreciated. Thanks.

P.S.
I'm aware of the .toBinary function, but I must create my own method for this one.

You got it done, the only thing you are missing is to REVERSE the binary string

int decimal=10;
    String binary = "";
    int remainder;

    while (decimal != 0)
    {
        remainder = decimal % 2;
        decimal /= 2;
        binary += remainder;
    }
    System.out.println(new StringBuilder(binary).reverse().toString());

yelds

1010 as result

You need to add remainder as binary = remainder + binary or reverse the string. And no need to use String.valueOf since it's already a string. And also consider the case when decimal is zero you need to send 0

  public String convertBinary(int decimal) {
    String binary = "";
    int remainder;
    if(decimal == 0) {
      return "0";
    }
    while (decimal != 0) {
      remainder = decimal % 2;
      decimal /= 2;
      binary = remainder + binary;
    }
    return binary;
  }

For a different twist, you can also do it like this.

• Initialize the result to an empty string.
• Using the AND & operator, mask off the low order bit. You could also use the remainder % operator here too.
• Convert the bit, 1 or 0, to a string and prepend it to the result string.
• Right shift the number (decimal) by 1 bit so the next bit can be checked. <br>
  while(!(decimal>>>1) == 0)
• continue the while loop until decimal == 0 (i.e. all 1 bits have been shifted out of the number).

public static String toBinary(int decimal) {
	// intialize the string
	String binary = Integer.toString(decimal&1);
    while ((decimal>>>=1) != 0) {
    	binary  =  Integer.toString(decimal&1)+ binary;
        // or binary = Integer.toString(decimal % 2) + binary;
    }
    return binary;
}


</details>



# 答案4
**得分**: 0

你尝试过这个吗？
```java
String test = Integer.toString(n, 2); //n是要转换的数字，2表示进制（二进制）。如果你想要一个八进制数，只需将进制改为8。

但是使用这个方法会在你想要将那个String重新转换回数字时引发问题，就像这样：

int number = Integer.parseInt(test); //这会导致NumberFormatException（数字格式异常）。

为了避免这个错误，必须将test的每个字符转换为一个char：

char digit = test.charAt(i);

最后，再将每个字符转换回一个单独的String：

String converted = String.valueOf(digit);

现在你可以将这个新的String转换为一个int。

String test = Integer.toString(n, 2); //n is the number to convert and 2 the base (binary). In case you want an octal number, for example, just change the base to 8.

int number = Integer.parseInt(test); //This will cause NumberFormatException.

char digit = test.charAt(i);

String converted = String.valueof(digit);