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I'm a noob when it comes to DownValues, UpValues, OwnValues, etc. but I figured I would give it a shot in my code because I think it would be beneficial for my work.

What I'm wondering is is it possible to set a DownValue for a compiled function? I've searched the docs and haven't come up with anything about DownValue's for compiled functions. Consider the following example

ClearAll[f, cf]
f[x_] := (x^2 + x)/x
cf[0.] = 1.;
cf = Compile[{x}, Evaluate[f[x]], CompilationTarget -> "C"];
cf[0.]

(*output*)
Indeterminate

The uncompiled function simplifies to an expression that has a clearly defined value at x=0, but the compiled function doesn't know that, so I'm trying to set a downvalue for the compiled function at the apparent singularity to rid this effect. This is only a simple example, the function that's relevant to me is much more complicated unfortunately, but has the same apparent singularity at the origin. If I could set a downvalue for the compiled function, that would be tremendously helpful.

Maybe something with OwnValues would work better here?

edit:

Ah, My ignorance on the types of values is showing. I see setting cf[0.]=1 sets the DownValue for the symbol cf, while setting cf = Compile... sets the OwnValue for cf. So, one possible solution to this would be

ClearAll[f, cf]
f[x_] := (x^2 + x)/x
compiledf =  Compile[{x}, Evaluate[f[x]], CompilationTarget -> "C"];
cf[0.] = 1.;
cf[a_] := compiledf[a];
cf[0.]

(*output*)
1.

But this is a bit of an ad hoc solution. Also the definition of compiledf needs to follow the definition of cf since:

DownValues[cf]

(*output*)
(*{HoldPattern[cf[0.]] :> 1., HoldPattern[cf[a_]] :> compiledf[a]}*)

So compiledf would also need to be passed to a subkernel if cf is ever passed to one, which is a bit inconvenient.

Maybe someone else has a better solution?

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ClearAll[f, cf]
f[x_] := (x^2 + x)/x

cf[0. | 0] = 1.;
With[{comp = Compile[{x}, Evaluate[f[x]]]},
  cf[a_] := comp[a]
]

cf[0.]  (* 1. *)
cf[0]   (* 1. *)
cf[2]   (* 3. *)
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  • $\begingroup$ awesome. Thank you! $\endgroup$ – shanedrum Jan 11 at 8:41

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