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floor function problem

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Old   May 24, 2014, 15:34
Default floor function problem
  #1
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Hi,

Trying to implement the following code snippet in a Ueqns routine:

Code:
volScalarField Ad
   (
        "Ad",
        floor( min( scalar(1.0), mag(Ua)/limitV ) )
    );
and I get the following error at compile,

Code:
UEqns.H:9:48: error: cannot convert ‘Foam::tmp<Foam::GeometricField<double, Foam::fvPatchField, Foam::volMesh> >’ to ‘double’ for argument ‘1’ to ‘double floor(double)’
The expression min( scalar(1.0), mag(Ua)/limitV ) compiles properly but it is refused by floor() which expects a double. I would have expected it to be converted automatically but it is not the case. Using the float version floorf() returns the same error with float instead of double.

Any ideas how it can be fixed?

Thanks
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Old   May 24, 2014, 18:34
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  #2
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Posting the solution just in case it comes in handy for someone else.

It seems floor() does not map over an entire array like min() does. The following modification compiled without any problems. Still have to test the code to see if I get the correct anticipated result.

Code:
volScalarField Ad
(
   "Ad",
   //floor( min( scalar(1.0), mag(Ua)/limitV ) )
   min( scalar(1.0), mag(Ua)/limitV )
 );

 forAll ( mesh.C(), celli )
 {
    Ad[celli] = floor( Ad[celli] );
 }
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