在Java中将两个浮点数与舍入进行比较（Big Java Ex 5.3）

Question

I've read a bunch of questions on rounding numbers to n amount of decimal places and then I've found out using BigDecimal is pretty good However I'm having a bit of trouble with the following exercise in my book, Consider the following input and output

Enter two floating-point numbers:
2.0
1.99998

//output
They are the same when rounded to two decimal places.
They differ by less than 0.01.

Enter two floating-point numbers:
0.999
0.991

//output
They are different when rounded to two decimal places.
They differ by less than 0.01.

MY CODE

public class test{
    public static void main(String [] args ){
        Scanner getNum = new Scanner(System.in);
        Scanner getNumTwo = new Scanner(System.in);

        BigDecimal a = getNum.nextBigDecimal();
        float b = getNum.nextFloat();

        BigDecimal valBNew = new BigDecimal(b).setScale(2,BigDecimal.ROUND_HALF_UP);
        BigDecimal valANew = a;
        System.out.println(valBNew); // 2.00
        System.out.println(a); //2.0
        if (valBNew == a){
            System.out.println("They are the same when rounded to two decimal places");
        }
        else{
            System.out.println("They are different when rounded to two decimal places");
        }

    }
}

Heres my attempt. I'm having trouble using two floats and then using big decimal. And then i'm having issues with my if statement since 2.0 isnt the same as 2.00 .

Whats my best option here? I'd like to get the rounding part working then the difference will be easy.

Answer 1

From the BigDecimal documentation .

public int compareTo(BigDecimal val)

Compares this BigDecimal with the specified BigDecimal . Two BigDecimal objects that are equal in value but have a different scale (like 2.0 and 2.00) are considered equal by this method.

Whereas, for .equals

public boolean equals(Object x)

Compares this BigDecimal with the specified Object for equality. Unlike compareTo , this method considers two BigDecimal objects equal only if they are equal in value and scale (thus 2.0 is not equal to 2.00 when compared by this method).

Therefore, this is correct

if (valBNew.compareTo(a) == 0) {

}

Answer 2

If a and b are both floating point types, then

(long)(a * 100 + 0.5) == (long)(b * 100 + 0.5)

is an adequate test for 2-decimal-place-rounding equality, subject to your not overflowing the long type, and your being able to tolerate the limitations of binary floating point. The 0.5 constants implement the German rounding.

Answer 3

Based on @cricket_007 -s answer, this is how a clean solution would look like:

public class Test {
    public static void main(String [] args ) {
        try (Scanner getNum = new Scanner(System.in)) { // try-with-resources, e.g. it auto-closes the scanner
            double a = Double.parseDouble(getNum.next()); // parsing a number from a line
            double b = Double.parseDouble(getNum.next());

            BigDecimal valBNew = new BigDecimal(b); // 2.00, no rounding
            BigDecimal valANew = new BigDecimal(a); // 2.0
            if (valBNew.compareTo(valANew) == 0) { // this is the magic
                System.out.println("They are the same");
            } else {
                System.out.println("They are different");
            }
        }
    }
}