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|   | ScalingFunctionFunctor ()=default | 
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|   | ScalingFunctionFunctor (double L, Level n, Translation l, int k) | 
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| double  | operator() (const coord_1d &r) const final | 
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| virtual Level  | special_level () const final | 
|   | Override this to change the minimum level of refinement at special points (default is 6)  
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| virtual std::vector< Vector< double, 1 > >  | special_points () const final | 
|   | Override this to return list of special points to be refined more deeply.  
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|   | FunctionFunctorInterface () | 
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| virtual  | ~FunctionFunctorInterface () | 
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| virtual coeffT  | coeff (const keyT &) const | 
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| virtual void  | operator() (const Vector< double *, 1 > &xvals, double *fvals, int npts) const | 
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| virtual void  | operator() (const Vector< double *, 2 > &xvals, double *fvals, int npts) const | 
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| virtual void  | operator() (const Vector< double *, 3 > &xvals, double *fvals, int npts) const | 
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| virtual void  | operator() (const Vector< double *, 4 > &xvals, double *fvals, int npts) const | 
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| virtual void  | operator() (const Vector< double *, 5 > &xvals, double *fvals, int npts) const | 
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| virtual void  | operator() (const Vector< double *, 6 > &xvals, double *fvals, int npts) const | 
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| virtual double  | operator() (const Vector< double, NDIM > &x) const=0 | 
|   | You should implement this to return f(x)  
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| virtual bool  | provides_coeff () const | 
|   | does this functor directly provide sum coefficients? or only function values?  
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| virtual bool  | screened (const Vector< double, NDIM > &c1, const Vector< double, NDIM > &c2) const | 
|   | Can we screen this function based on the bounding box information?  
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| void  | set_length_scale (double lo) | 
|   | adapt the special level to resolve the smallest length scale  
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| virtual bool  | supports_vectorized () const | 
|   | Does the interface support a vectorized operator()?  
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| virtual coeffT  | values (const keyT &key, const Tensor< double > &tensor) const | 
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