add method

Big add(dynamic y)

Return a new Big whose value is the value of this Big plus the value of Big y.

Implementation

Big add(dynamic y) {
  var yBig = Big(y);
  int e, k;
  List<int> t;
  Big x = this;

  // Signs differ?
  if (x.s != yBig.s) {
    yBig.s = -yBig.s;
    return x.sub(yBig);
  }

  var xe = x.e, xc = x.c, ye = yBig.e, yc = yBig.c;
  // Either zero?
  if (!xc.isElementIsValid(0) || !yc.isElementIsValid(0)) {
    if (!yc.isElementIsValid(0)) {
      if (xc.isElementIsValid(0)) {
        yBig = Big(x);
      } else {
        yBig.s = x.s;
      }
    }
    return yBig;
  }
  xc = [...xc];

  // Prepend zeros to equalise exponents.
  e = xe - ye;
  if (e != 0) {
    if (e > 0) {
      ye = xe;
      t = yc;
    } else {
      e = -e;
      t = xc;
    }

    t.reverse();

    for (; e-- > 0;) {
      t.add(0);
    }
    t.reverse();
  }

  // Point xc to the longer array.
  if (xc.length - yc.length < 0) {
    t = yc;
    yc = xc;
    xc = t;
  }

  e = yc.length;

  // Only start adding at yc.length - 1 as the further digits of xc can be left as they are.
  for (k = 0; e > 0; xc[e] %= 10) {
    k = (xc[--e] = xc[e] + yc[e] + k) ~/ 10 | 0;
  }

  // No need to check for zero, as +x + +y != 0 && -x + -y != 0

  if (k != 0) {
    unShift(xc, k);
    ++ye;
  }

  // Remove trailing zeros.
  for (e = xc.length; xc[--e] == 0;) {
    xc.removeLast();
  }

  yBig.c = xc;
  yBig.e = ye;

  return yBig;
}