BigDecimal & BigInteger — exact arithmetic

Why floating-point fails for money, and how BigDecimal solves it.

The floating-point problem

Binary floating-point (double, float) cannot represent most decimal fractions exactly. The classic demonstration:

System.out.println(0.1 + 0.2);   // 0.30000000000000004

For financial calculations, exam scores, or anywhere exactness matters, use BigDecimal:

BigDecimal a = new BigDecimal("0.1");   // from String — EXACT
BigDecimal b = new BigDecimal("0.2");   // from String — EXACT
BigDecimal c = a.add(b);               // 0.3 — exactly

Construction from a double re-introduces the imprecision you were trying to avoid:

new BigDecimal(0.1)  // → 0.1000000000000000055511151231257827021181583404541015625

Always construct BigDecimal from a String.

Arithmetic operations — all return new BigDecimal instances:

c.subtract(b);                           // subtraction
c.multiply(new BigDecimal("2"));         // multiplication
c.divide(new BigDecimal("3"), 2, RoundingMode.HALF_UP); // scale 2 decimal places
c.setScale(2, RoundingMode.HALF_UP);     // round to 2 decimal places

Comparison: use compareTo, not equals. equals also compares scale: new BigDecimal("2.0").equals(new BigDecimal("2.00")) is false. compareTo returns 0 for equal value:

new BigDecimal("2.0").compareTo(new BigDecimal("2.00")) == 0  // true

BigInteger handles arbitrarily large integers without overflow:

BigInteger factorial100 = BigInteger.ONE;
for (int i = 1; i <= 100; i++) factorial100 = factorial100.multiply(BigInteger.valueOf(i));