зеркало из https://github.com/mozilla/gecko-dev.git
138 строки
5.2 KiB
C++
138 строки
5.2 KiB
C++
/* -*- Mode: C++; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
|
|
/* vim: set ts=8 sts=2 et sw=2 tw=80: */
|
|
/* This Source Code Form is subject to the terms of the Mozilla Public
|
|
* License, v. 2.0. If a copy of the MPL was not distributed with this
|
|
* file, You can obtain one at http://mozilla.org/MPL/2.0/. */
|
|
|
|
|
|
#include <math.h>
|
|
#include "nsGeoGridFuzzer.h"
|
|
#include "nsGeoPosition.h"
|
|
|
|
|
|
#ifdef MOZ_APPROX_LOCATION
|
|
|
|
/* The following constants are taken from the World Geodetic System 1984 (WGS84)
|
|
* reference model for the earth ellipsoid [1]. The values in the model are
|
|
* an accepted standard for GPS and other navigational systems.
|
|
*
|
|
* [1] http://www.oosa.unvienna.org/pdf/icg/2012/template/WGS_84.pdf
|
|
*/
|
|
#define WGS84_a (6378137.0) // equitorial axis
|
|
#define WGS84_b (6356752.314245179) // polar axis (a * (1-f))
|
|
#define WGS84_f (1.0/298.257223563) // inverse flattening
|
|
#define WGS84_EPSILON (5.72957795e-9) // 1e-10 radians in degrees
|
|
#define sq(f) ((f) * (f))
|
|
#define sign(f) (((f) < 0) ? -1 : 1)
|
|
|
|
/* if you have an ellipsoid with semi-major axis A and semi-minor axis B, the
|
|
* radius at angle phi along the semi-major axis can be calculated with this
|
|
* formula. by using the WGS84 values for A and B, we calculate the radius of
|
|
* earth, given the angle of latitude, phi.*/
|
|
#define LON_RADIUS(phi) (sqrt((sq(sq(WGS84_a) * cos(phi)) + sq(sq(WGS84_b) * sin(phi))) / \
|
|
(sq(WGS84_a * cos(phi)) + sq(WGS84_b * sin(phi)))))
|
|
/* the radius of earth changes as a function of latitude, to simplify I am
|
|
* assuming the fixed radius of the earth halfway between the poles and the
|
|
* equator. this is calculated from LON_RADIUS(M_PI/4), or the radius at
|
|
* 45 degrees N.*/
|
|
#define LAT_RADIUS (6367489.543863)
|
|
|
|
/* This function figures out the latitudinal grid square that the given
|
|
* latitude coordinate falls into and then returns the latitudinal center of
|
|
* that grid square. It handles the proper wrapping at the poles +/- 90
|
|
* (e.g. +95 wraps to +85 and -95 wraps to -85) */
|
|
static double GridAlgorithmLat(int32_t aDistMeters, double aLatDeg)
|
|
{
|
|
/* switch to radians */
|
|
double phi = (aLatDeg * M_PI) / 180;
|
|
|
|
/* properly wrap the latitude */
|
|
phi = atan(sin(phi) / fabs(cos(phi)));
|
|
|
|
/* calculate grid size in radians */
|
|
double gridSizeRad = aDistMeters / LAT_RADIUS;
|
|
|
|
/* find the southern edge, in radians, of the grid cell, then add half of a
|
|
* grid cell to find the center latitude in radians */
|
|
double gridCenterPhi = gridSizeRad * floor(phi / gridSizeRad) + gridSizeRad / 2;
|
|
|
|
/* properly wrap it and return it in degrees */
|
|
return atan(sin(gridCenterPhi) / fabs(cos(gridCenterPhi))) * (180.0 / M_PI);
|
|
}
|
|
|
|
/* This function figures out the longitudinal grid square that the given longitude
|
|
* coordinate falls into and then returns the longitudinal center of that grid
|
|
* square. It handles the proper wrapping at +/- 180 (e.g. +185 wraps to -175
|
|
* and -185 wraps to +175) */
|
|
static double GridAlgorithmLon(int32_t aDistMeters, double aLatDeg, double aLonDeg)
|
|
{
|
|
/* switch to radians */
|
|
double phi = (aLatDeg * M_PI) / 180;
|
|
double theta = (aLonDeg * M_PI) / 180;
|
|
|
|
/* properly wrap the lat/lon */
|
|
phi = atan(sin(phi) / fabs(cos(phi)));
|
|
theta = atan2(sin(theta), cos(theta));
|
|
|
|
/* calculate grid size in radians */
|
|
double gridSizeRad = aDistMeters / LON_RADIUS(phi);
|
|
|
|
/* find the western edge, in radians, of the grid cell, then add half of a
|
|
* grid cell to find the center longitude in radians */
|
|
double gridCenterTheta = gridSizeRad * floor(theta / gridSizeRad) + gridSizeRad / 2;
|
|
|
|
/* properly wrap it and return it in degrees */
|
|
return atan2(sin(gridCenterTheta), cos(gridCenterTheta)) * (180.0 / M_PI);
|
|
}
|
|
|
|
/* This function takes the grid size and the graticule coordinates of a
|
|
* location and calculates which grid cell the coordinates fall within and
|
|
* then returns the coordinates of the geographical center of the grid square.
|
|
*/
|
|
static void CalculateGridCoords(int32_t aDistKm, double& aLatDeg, double& aLonDeg)
|
|
{
|
|
// a grid size of 0 is the same as precise
|
|
if (aDistKm == 0) {
|
|
return;
|
|
}
|
|
aLonDeg = GridAlgorithmLon(aDistKm * 1000, aLatDeg, aLonDeg);
|
|
aLatDeg = GridAlgorithmLat(aDistKm * 1000, aLatDeg);
|
|
}
|
|
|
|
already_AddRefed<nsIDOMGeoPosition>
|
|
nsGeoGridFuzzer::FuzzLocation(const GeolocationSetting & aSetting,
|
|
nsIDOMGeoPosition * aPosition)
|
|
{
|
|
if (!aPosition) {
|
|
return nullptr;
|
|
}
|
|
|
|
nsCOMPtr<nsIDOMGeoPositionCoords> coords;
|
|
nsresult rv = aPosition->GetCoords(getter_AddRefs(coords));
|
|
NS_ENSURE_SUCCESS(rv, nullptr);
|
|
if (!coords) {
|
|
return nullptr;
|
|
}
|
|
|
|
double lat = 0.0, lon = 0.0;
|
|
coords->GetLatitude(&lat);
|
|
coords->GetLongitude(&lon);
|
|
|
|
// adjust lat/lon to be the center of the grid square
|
|
CalculateGridCoords(aSetting.GetApproxDistance(), lat, lon);
|
|
GPSLOG("approximate location with delta %d is %f, %f",
|
|
aSetting.GetApproxDistance(), lat, lon);
|
|
|
|
// reusing the timestamp
|
|
DOMTimeStamp ts;
|
|
rv = aPosition->GetTimestamp(&ts);
|
|
NS_ENSURE_SUCCESS(rv, nullptr);
|
|
|
|
// return a position at sea level, N heading, 0 speed, 0 error.
|
|
RefPtr<nsGeoPosition> pos = new nsGeoPosition(lat, lon, 0.0, 0.0,
|
|
0.0, 0.0, 0.0, ts);
|
|
return pos.forget();
|
|
}
|
|
|
|
#endif
|