Hi,
A TF1 can only use free standing function (it would not know which object to
call a member function on).
Before 3.05/06 you needed to declare a free standing function with the exact
signature described in the TF1 documentation.
With 3.05/06 you can access any free standing function or static data member
functions, using a syntax like:
TF1 *fitFcn = new
TF1("fitFcn","func_fit::func_val(x,y,[0])",-600.,-1.,fclass.get_nparam());
with func_val being 'static double func_fit::func_val(double arg1, double
arg2, double param)'
Cheers,
Philippe
-----Original Message-----
From: owner-roottalk@pcroot.cern.ch
[mailto:owner-roottalk@pcroot.cern.ch]On Behalf Of Dr. John Krane
Sent: Tuesday, July 22, 2003 9:06 AM
To: roottalk@pcroot.cern.ch
Subject: [ROOT] fitting with a member function
Hi,
I'm trying to use weighted histograms to fit a distribution. Because I
want to be able to change the histos that I'm using as components, I
made a little class that holds an array of TH1F. A member function
provides the weighted sum of these histograms, with the weights being
the fit parameters and the interface being as required by TF1. So I
tried to make a TF1 for use with the ->Fit method, like this:
cerr<< "Assign the histos that contribute to fit"<<endl;
func_fit fclass;
fclass.add_param(d1or); //first histo
fclass.add_param(d1crl); //second histo...etc.
TF1 *fitFcn = new
TF1("fitFcn",fclass.func_val,-600.,-1.,fclass.get_nparam());
Well the compiler within ROOT doesn't like that, so I tried making a
pointer-to-member-function and giving that to root, but I didn't get
anywhere. Does anybody know how to provide a member-function to TF1? I
think my class solution is quite clean for what I'd like to do, but I
need to get around this barrier...
- John
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