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Fixed misleading comments
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@ -444,20 +444,20 @@ inline VFPEvaluation interpolate(
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/**
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* This basically models interpolate(VFPProdTable::array_type, ...)
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* which performs 5D interpolation, but here for the 2D case only
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*/
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inline VFPEvaluation interpolate(
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const VFPInjTable::array_type& array,
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const InterpData& flo_i,
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const InterpData& thp_i) {
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//Values and derivatives in a 5D hypercube
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//Values and derivatives in a 2D plane
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VFPEvaluation nn[2][2];
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//Pick out nearest neighbors (nn) to our evaluation point
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//This is not really required, but performance-wise it may pay off, since the 32-elements
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//we copy to (nn) will fit better in cache than the full original table for the
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//interpolation below.
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//The following ladder of for loops will presumably be unrolled by a reasonable compiler.
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for (int t=0; t<=1; ++t) {
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for (int f=0; f<=1; ++f) {
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@ -489,16 +489,13 @@ inline VFPEvaluation interpolate(
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double t1, t2; //interpolation variables, so that t1 = (1-t) and t2 = t.
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// Remove dimensions one by one
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// Example: going from 3D to 2D to 1D, we start by interpolating along
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// the z axis first, leaving a 2D problem. Then interpolating along the y
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// axis, leaving a 1D, problem, etc.
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// Go from 2D to 1D
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t2 = flo_i.factor_;
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t1 = (1.0-t2);
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for (int t=0; t<=1; ++t) {
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nn[t][0] = t1*nn[t][0] + t2*nn[t][1];
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}
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nn[0][0] = t1*nn[0][0] + t2*nn[0][1];
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nn[1][0] = t1*nn[1][0] + t2*nn[1][1];
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// Go from line to point on line
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t2 = thp_i.factor_;
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t1 = (1.0-t2);
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nn[0][0] = t1*nn[0][0] + t2*nn[1][0];
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