Convert double to BigDecimal and set BigDecimal Precision

In Java, I want to take a double value and convert it to a BigDecimal and print out its String value to a certain precision.

import java.math.BigDecimal;
public class Main {
    public static void main(String[] args) {
        double d=-.00012;
        System.out.println(d+""); //This prints -1.2E-4

        double c=47.48000;
        BigDecimal b = new BigDecimal(c);
        System.out.println(b.toString()); 
        //This prints 47.47999999999999687361196265555918216705322265625 
    }
}

It prints this huge thing:

47.47999999999999687361196265555918216705322265625

and not

47.48

The reason I'm doing the BigDecimal conversion is sometimes the double value will contain a lot of decimal places (i.e. -.000012) and the when converting the double to a String will produce scientific notation -1.2E-4. I want to store the String value in non-scientific notation.

I want to have BigDecimal always have two units of precision like this: "47.48". Can BigDecimal restrict precision on conversion to string?

The reason of such behaviour is that the string that is printed is the exact value - probably not what you expected, but that's the real value stored in memory - it's just a limitation of floating point representation.

According to javadoc, BigDecimal(double val) constructor behaviour can be unexpected if you don't take into consideration this limitation:

The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.

So in your case, instead of using

double val = 77.48;
new BigDecimal(val);

use

BigDecimal.valueOf(val);

Value that is returned by BigDecimal.valueOf is equal to that resulting from invocation of Double.toString(double).

It prints 47.48000 if you use another MathContext:

BigDecimal b = new BigDecimal(d, MathContext.DECIMAL64);

Just pick the context you need.

You want to try String.format("%f", d), which will print your double in decimal notation. Don't use BigDecimal at all.

Regarding the precision issue: You are first storing 47.48 in the double c, then making a new BigDecimal from that double. The loss of precision is in assigning to c. You could do

BigDecimal b = new BigDecimal("47.48")

to avoid losing any precision.