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forgejo/modules/avatar/identicon/polygon.go
Gusted ff2fd08228
Simplify parameter types (#18006)
Remove repeated type declarations in function definitions.
2021-12-20 04:41:31 +00:00

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// Copyright 2021 The Gitea Authors. All rights reserved.
// Use of this source code is governed by a MIT-style
// license that can be found in the LICENSE file.
// Copied and modified from https://github.com/issue9/identicon/ (MIT License)
package identicon
var (
// cos(0),cos(90),cos(180),cos(270)
cos = []int{1, 0, -1, 0}
// sin(0),sin(90),sin(180),sin(270)
sin = []int{0, 1, 0, -1}
)
// rotate the points by center point (x,y)
// angle: [0,1,2,3] means [090180270] degree
func rotate(points []int, x, y, angle int) {
// the angle is only used internally, and it has been guaranteed to be 0/1/2/3, so we do not check it again
for i := 0; i < len(points); i += 2 {
px, py := points[i]-x, points[i+1]-y
points[i] = px*cos[angle] - py*sin[angle] + x
points[i+1] = px*sin[angle] + py*cos[angle] + y
}
}
// check whether the point is inside the polygon (defined by the points)
// the first and the last point must be the same
func pointInPolygon(x, y int, polygonPoints []int) bool {
if len(polygonPoints) < 8 { // a valid polygon must have more than 2 points
return false
}
// reference: nonzero winding rule, https://en.wikipedia.org/wiki/Nonzero-rule
// split the plane into two by the check point horizontally:
// y>0includes (x>0 && y==0)
// y<0includes (x<0 && y==0)
//
// then scan every point in the polygon.
//
// if current point and previous point are in different planes (eg: curY>0 && prevY<0),
// check the clock-direction from previous point to current point (use check point as origin).
// if the direction is clockwise, then r++, otherwise then r--
// finally, if 2==abs(r), then the check point is inside the polygon
r := 0
prevX, prevY := polygonPoints[0], polygonPoints[1]
prev := (prevY > y) || ((prevX > x) && (prevY == y))
for i := 2; i < len(polygonPoints); i += 2 {
currX, currY := polygonPoints[i], polygonPoints[i+1]
curr := (currY > y) || ((currX > x) && (currY == y))
if curr == prev {
prevX, prevY = currX, currY
continue
}
if mul := (prevX-x)*(currY-y) - (currX-x)*(prevY-y); mul >= 0 {
r++
} else { // mul < 0
r--
}
prevX, prevY = currX, currY
prev = curr
}
return r == 2 || r == -2
}