# noise2D method Null safety

double noise2D(
1. double xin,
2. double yin
)

## Implementation

``````double noise2D(double xin, double yin) {
double n0, n1, n2; // Noise contributions from the three corners
// Skew the input space to determine which simplex cell we're in
final s = (xin + yin) * _F2; // Hairy factor for 2D
final i = (xin + s).floor();
final j = (yin + s).floor();
final t = (i + j) * _G2;
final X0 = i - t; // Unskew the cell origin back to (x,y) space
final Y0 = j - t;
final x0 = xin - X0; // The x,y distances from the cell origin
final y0 = yin - Y0;
// For the 2D case, the simplex shape is an equilateral triangle.
// Determine which simplex we are in.
int i1, j1; // Offsets for second (middle) corner of simplex in (i,j) coords
if (x0 > y0) {
i1 = 1;
j1 = 0;
} // lower triangle, XY order: (0,0)->(1,0)->(1,1)
else {
i1 = 0;
j1 = 1;
} // upper triangle, YX order: (0,0)->(0,1)->(1,1)
// A step of (1,0) in (i,j) means a step of (1-c,-c) in (x,y), and
// a step of (0,1) in (i,j) means a step of (-c,1-c) in (x,y), where
// c = (3-sqrt(3))/6
final x1 =
x0 - i1 + _G2; // Offsets for middle corner in (x,y) unskewed coords
final y1 = y0 - j1 + _G2;
final x2 = x0 -
1.0 +
2.0 * _G2; // Offsets for last corner in (x,y) unskewed coords
final y2 = y0 - 1.0 + 2.0 * _G2;
// Work out the hashed gradient indices of the three simplex corners
final ii = i & 255;
final jj = j & 255;
final gi0 = _permMod12[ii + _perm[jj]];
final gi1 = _permMod12[ii + i1 + _perm[jj + j1]];
final gi2 = _permMod12[ii + 1 + _perm[jj + 1]];
// Calculate the contribution from the three corners
var t0 = 0.5 - x0 * x0 - y0 * y0;
if (t0 < 0) {
n0 = 0.0;
} else {
t0 *= t0;
n0 = t0 *
t0 *
}
var t1 = 0.5 - x1 * x1 - y1 * y1;
if (t1 < 0) {
n1 = 0.0;
} else {
t1 *= t1;
n1 = t1 * t1 * _dot2(_grad3[gi1], x1, y1);
}
var t2 = 0.5 - x2 * x2 - y2 * y2;
if (t2 < 0) {
n2 = 0.0;
} else {
t2 *= t2;
n2 = t2 * t2 * _dot2(_grad3[gi2], x2, y2);
}
// Add contributions from each corner to get the final noise value.
// The result is scaled to return values in the interval [-1,1].
return 70.0 * (n0 + n1 + n2);
}``````