h1or4dme8/osu-lazer
1
0
Fork 0
mirror of https://github.com/ppy/osu.git synced 2024-11-13 16:13:34 +08:00
Code Issues Releases Wiki Activity

Fix InverseDifficultyRange() not working correctly in both directions

Browse Source
This commit is contained in:
Bartłomiej Dach 2023-12-14 19:30:56 +01:00
parent fd1c72bf74
commit c0e68df20f
No known key found for this signature in database
1 changed files with 9 additions and 7 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

16
osu.Game/Beatmaps/IBeatmapDifficultyInfo.cs
View File

@ -1,6 +1,8 @@
// Copyright (c) ppy Pty Ltd <contact@ppy.sh>. Licensed under the MIT Licence.
// See the LICENCE file in the repository root for full licence text.
using System;
namespace osu.Game.Beatmaps
{
/// <summary>
@ -98,15 +100,15 @@ namespace osu.Game.Beatmaps
/// Maps a value returned by the function above back to the difficulty that produced it.
/// </summary>
/// <param name="difficultyValue">The difficulty-dependent value to be unmapped.</param>
/// <param name="min">Minimum of the resulting range which will be achieved by a difficulty value of 0.</param>
/// <param name="mid">Midpoint of the resulting range which will be achieved by a difficulty value of 5.</param>
/// <param name="max">Maximum of the resulting range which will be achieved by a difficulty value of 10.</param>
/// <param name="diff0">Minimum of the resulting range which will be achieved by a difficulty value of 0.</param>
/// <param name="diff5">Midpoint of the resulting range which will be achieved by a difficulty value of 5.</param>
/// <param name="diff10">Maximum of the resulting range which will be achieved by a difficulty value of 10.</param>
/// <returns>Value to which the difficulty value maps in the specified range.</returns>
static double InverseDifficultyRange(double difficultyValue, double min, double mid, double max)
static double InverseDifficultyRange(double difficultyValue, double diff0, double diff5, double diff10)
{
return difficultyValue >= mid
? (difficultyValue - mid) / (max - mid) * 5 + 5
: (difficultyValue - mid) / (mid - min) * 5 + 5;
return Math.Sign(difficultyValue - diff5) == Math.Sign(diff10 - diff5)
? (difficultyValue - diff5) / (diff10 - diff5) * 5 + 5
: (difficultyValue - diff5) / (diff5 - diff0) * 5 + 5;
}
}
}