isCloseTo method

@useResult bool isCloseTo(double other, { double relativeTolerance = 1e-9, double absoluteTolerance = 1e-12, })

Returns true if this value is within tolerance of other.

Direct == on doubles is unreliable: 0.1 + 0.2 == 0.3 is false because binary floating point cannot represent those decimals exactly. This compares with a tolerance instead, so accumulated rounding error does not register as inequality.

The tolerance is the larger of absoluteTolerance (a fixed floor, which makes comparisons against 0.0 meaningful — a relative tolerance alone collapses to zero there) and relativeTolerance scaled by the magnitude of the inputs (so the same call works for both tiny and very large numbers). Defaults follow the common 1e-9 convention.

NaN is never close to anything, including another NaN. Two infinities of the same sign are close; opposite signs are not.

Example:

(0.1 + 0.2).isCloseTo(0.3); // true
1.0.isCloseTo(1.5);         // false
0.0.isCloseTo(1e-12);       // true (absolute floor)

Audited: 2026-06-12 11:26 EDT

Implementation

@useResult
bool isCloseTo(
  double other, {
  double relativeTolerance = 1e-9,
  double absoluteTolerance = 1e-12,
}) {
  // Exact match (covers same-sign infinity) without computing a difference
  // that would be NaN for inf - inf.
  if (this == other) {
    return true;
  }

  // NaN and any remaining infinity case cannot be "close": a finite tolerance
  // can never bridge them.
  if (isNaN || other.isNaN || isInfinite || other.isInfinite) {
    return false;
  }

  final double diff = (this - other).abs();
  final double scaled = relativeTolerance * max(abs(), other.abs());
  return diff <= max(absoluteTolerance, scaled);
}