BigDecimal.divide() yields extra digit of precision when called with a MathContext

Question

Given the simple program

import java.math.*;
import static java.math.BigDecimal.ONE;
import static java.lang.System.out;

public static void main(String[] args) {
    StringBuffer ruler = new StringBuffer("  ");
    for (int i = 0; i < 5; i++) {
         ruler.append("1234567890");
    }

    out.println(ONE.divide(new BigDecimal(47), 50, RoundingMode.HALF_UP));
    out.println(ONE.divide(new BigDecimal(47), new MathContext(50, RoundingMode.HALF_UP)));
    out.println(ruler);
    out.println(ONE.divide(new BigDecimal(6), 5, RoundingMode.HALF_UP));
    out.println(ONE.divide(new BigDecimal(6), new MathContext(5, RoundingMode.HALF_UP)));
}

This is the output:

0.02127659574468085106382978723404255319148936170213
0.021276595744680851063829787234042553191489361702128
  12345678901234567890123456789012345678901234567890
0.16667
0.16667

I would expect the second line of output to be the same as the first line. Is that a bug, or have I misinterpreted the BigDecimal documentation?

JVM version:

$ java -version
java version "1.7.0_45"
Java(TM) SE Runtime Environment (build 1.7.0_45-b18)
Java HotSpot(TM) 64-Bit Server VM (build 24.45-b08, mixed mode)

Answer 1

You have confused scale (total number of decimal places) with precision (number of significant digits). For numbers between -1 and 1, the precision does not count any zeroes between the decimal point and the non-zero decimal places, but the scale does.

The second argument to BigDecimal.divide is a scale. So you get 50 decimal places for your first output.

The argument to the MathContext constructor is a precision. So for your second output, you get 50 significant decimal places, plus one additional zero between the decimal point and the 2.

  First decimal place (start counting scale from here)
  ↓
0.02127659574468085106382978723
   ↑
   First significant digit (start counting precision from here)