text: /*
   * Copyright 2012 Google Inc.
   *
   * Use of this source code is governed by a BSD-style license that can be
   * found in the LICENSE file.
   */
  
  // inline utilities
  /* Returns 0 if negative, 1 if zero, 2 if positive
  */
  inline int side(double x) {
      return (x > 0) + (x >= 0);
  }
  
  /* Returns 1 if negative, 2 if zero, 4 if positive
  */
  inline int sideBit(double x) {
      return 1 << side(x);
  }
  
  /* Given the set [0, 1, 2, 3], and two of the four members, compute an XOR mask
     that computes the other two. Note that:
  
     one ^ two == 3 for (0, 3), (1, 2)
     one ^ two <  3 for (0, 1), (0, 2), (1, 3), (2, 3)
     3 - (one ^ two) is either 0, 1, or 2
     1 >> 3 - (one ^ two) is either 0 or 1
  thus:
     returned == 2 for (0, 3), (1, 2)
     returned == 3 for (0, 1), (0, 2), (1, 3), (2, 3)
  given that:
     (0, 3) ^ 2 -> (2, 1)  (1, 2) ^ 2 -> (3, 0)
     (0, 1) ^ 3 -> (3, 2)  (0, 2) ^ 3 -> (3, 1)  (1, 3) ^ 3 -> (2, 0)  (2, 3) ^ 3 -> (1, 0)
  */
  inline int other_two(int one, int two) {
      return 1 >> 3 - (one ^ two) ^ 3;
  }
  
  /* Returns -1 if negative, 0 if zero, 1 if positive
  */
  inline int sign(double x) {
      return (x > 0) - (x < 0);
  }
  
  inline double interp(double A, double B, double t) {
      return A + (B - A) * t;
  }
  