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RRExponentialFunction.m
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// RRUtils RRExponentialFunction.m
//
// Copyright © 2008-2010, Roy Ratcliffe, Pioneering Software, United Kingdom
// All rights reserved
//
// Permission is hereby granted, free of charge, to any person obtaining a copy
// of this software and associated documentation files (the "Software"), to deal
// in the Software without restriction, including without limitation the rights
// to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
// copies of the Software, and to permit persons to whom the Software is
// furnished to do so, subject to the following conditions:
//
// The above copyright notice and this permission notice shall be included in
// all copies or substantial portions of the Software.
//
// THE SOFTWARE IS PROVIDED "AS IS," WITHOUT WARRANTY OF ANY KIND, EITHER
// EXPRESSED OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
// MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NON-INFRINGEMENT. IN NO
// EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES
// OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE,
// ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
// DEALINGS IN THE SOFTWARE.
//
//------------------------------------------------------------------------------
#import "RRExponentialFunction.h"
#import <math.h> // for expf
void RRExponentialFunctionSetCoefficient(RRExponentialFunction *f, float c)
{
f->coefficient = c;
f->exponentOfMinusCoefficient = expf(-c);
f->oneOverOneMinusExponentOfMinusCoefficient = 1.0f/(1.0f - f->exponentOfMinusCoefficient);
// Computing "one over one minus exponent of minus coefficient" optimises
// for multiplication over division. Effectively, the exponential function
// divides by one minus the exponent of the coefficient. Deriving the
// reciprocal upfront changes the divide to multiply during
// evaluation. Multiply typically beats divide when it comes to performance.
}
float RRExponentialFunctionEvaluate(RRExponentialFunction *f, float x)
{
return 1.0f - ((expf(x * -f->coefficient) - f->exponentOfMinusCoefficient) * f->oneOverOneMinusExponentOfMinusCoefficient);
}