Re: name for histograms

From: Axel Naumann <>
Date: Wed, 21 Jun 2006 16:37:54 +0200


>> first of all, you should use pointers to histograms:
>> class MyClass : public TObject {
>> private:
>>   TH1D *hist;

> 'hist' is a 'value object' for MyClass. (NOT 'reference object')
> That's why I did not declare hist as a pointer.
> Is there any reason (from ROOT constraints) to use pointers ?

Not a ROOT reason, but a C++ one. Suppose you have a TH1D hist, not a pointer, and you create an object of type MyClass:

the compiler invokes the default constructor TH1D::TH1D() for hist, and then you revert all that did later, setting up hist properly (name, binning,...), so it becomes usable. That's a waste of CPU, and you might have to watch out with renaming histograms (ROOT doesn't notice that they're renamed, and doesn't update its list of known histograms - which can be searched by name).

Or you need to initialize hist as part of the initialization list of the constructor of your class, i.e.
MyClass::MyClass(const char* name):
  hist(name, Form("a hist in MyClass named %s", name), 100, 0., 1.) {...}
Often you don't know yet the binning, the name, etc at the time the constructor is called.

That's why I always prefer the pointer member, and a simple MyClass::MyClass(const char* name): hist(0) {...}

creating hist = new TH1D(...) when everything is known.

That's the generic advice. Without knowing your use case and code I cannot tell whether having a TH1D member makes sense or not in your application.

Cheers, Axel. Received on Wed Jun 21 2006 - 16:38:02 MEST

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