BigDecimal.divide () gives an extra precision digit when called using a MathContext

Question

Given a 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 conclusion:

0.02127659574468085106382978723404255319148936170213
0.021276595744680851063829787234042553191489361702128
  12345678901234567890123456789012345678901234567890
0.16667
0.16667

I expect the second line of output to be the same as the first line. Is this a mistake, or am I misinterpreting the documentation BigDecimal?

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 an intricate scale (total number of decimal places) with precision (the number of significant digits). For numbers from -1 to 1, precision does not take into account zeros between the decimal point and nonzero decimal places, but the scale does.

The second argument BigDecimal.divideis scale. This way you get 50 decimal places for your first exit.

MathContext . , 50 , 2.

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