二进制运算符“ *”的错误操作数类型如何将类型转换为int

Question

I'm having some problems with the following program.

    public static void main(String[] args) {

    Int x =new Int(3);
    int y= square() +twice() +once();
    System.out.println(y);
}

private int square (Int x)
{
    x= x*x;
    return x;
}

private int twice (Int x)
{
    x= 2*x;
    return x;
}

private int once (Int x)
{
    x= x;
    return x;
}

and the output for this program should be 45.

here is the Int class.

public class Int {
private int x;
public Int (int x)
{
    this.x=x;
}

the problem I have is in

  private int square (Int x)
  {
      x= x*x;
      return x;
  }

x=x x gives an error of bad operand types for binary operator ' '.first type Int, Second Type Int.

I know for '*' to work it need an int type, I tried to use Integer.parseInt(x), but it says, x is not a string.

can someone help me? what causes this problem and how to fix it.

Answer 1

The problem is simple: you define an Int type and expect it to be implicitly convertible to a primitive int but this has nothing to do with what your Int type is expected to be in your design, it could be called Foo instead that Int but that would be the same.

If you want to be able to get the int value wrapped inside an Int instance then you must provide your own way to do it, eg:

class Int {
  int x;

  public int intValue() { return x; }
}

so that you can do:

Int x = new Int(3);
int square = x.intValue() * x.intValue();

or, to avoid breaking encapsulation:

class Int
{
  int x;

  public Int(int x) { this.x = x; }

  public Int square() { return new Int(x*x); }
  /* or: int square() { return x*x; }*/
}

Answer 2

Your class Int looks (very lightly) like the Integer wrapper. If you really need a wrapper, use the very tested Java Integer , otherwise, use the primitive int . Remember, Java does not support operator overloading.