Copying a histogram in a function

[cmiller]

Hello,

Is there a recommended or proper way to pass a histogram to a function, set it to a copy of an existing histogram and have it retain the info? Currently the histogram I pass is filled properly but lost when the function ends. Here’s a MWE demonstrating the issue:

MWE.C

void FillHist(TH1F* hist)
{
    TH1F* temp = new TH1F("h1","h1",10,0,1);
    temp->Fill(0.5);
    hist=(TH1F*)temp->Clone();
    cout<<hist->GetEntries()<<endl;
}
void MWE()
{
    TH1F* hist = new TH1F("h","h",10,0,1);
    FillHist(hist);
    cout<<hist->GetEntries()<<endl;
}

Output:

1
0

In the example the histogram “hist” will have 1 entry at the end of FillHist, however the print statement at the end of MWE() will print 0. I assume this is due to the scope of hist not extending past the function after the clone. Is there a better way to copy temp into hist without losing the data?

I probably could set up FillHist to return a TH1F instead, but for my use case I’d like to pass and fill/combine 3 different histograms which is trickier to return

_ROOT Version:6.20.04
_Platform:CentOS
_Compiler:gcc-9.3.0

[Wile_E_Coyote]

void FillHist(TH1F* &hist)

[cmiller]

Thanks that works, Do you mind giving me a quick explanation as too why it works? The part I don’t understand is how this relates to the Clone(). For example if I don’t clone and just fill the histogram:

void FillHist(TH1F* hist)
{
    hist->Fill(0.5);
    cout<<hist->GetEntries()<<endl;
}
void MWE()
{
    TH1F* hist = new TH1F("h","h",10,0,1);
    FillHist(hist);
    cout<<hist->GetEntries()<<endl;
}

I get an output of 1 and 1. So why is the & not needed here while it is for the Clone version?

[Wile_E_Coyote]

Uncle Google → C++ reference to pointer

[cmiller]

So why does the second example work then? The google result suggests it should only work with the &. To quote the first google result: “But the problem comes when you modify the pointer inside the function. Instead of modifying the variable, you are only modifying a copy of the pointer and the original pointer remains unmodified.”

Isn’t that what I’m doing in the second example? Yet it still successfully modifies the original pointer

[Wile_E_Coyote]

In your “second example”, you directly modify the original histogram.

[cmiller]

I think I sort of see the difference. From what I understand it seems to not require the & if I don’t overwrite the histogram with a new object.
So switching the Clone to Add also solves my original question.

hist->Add(temp,1);

instead of

hist = (TH1F*)temp->Clone();

I think my remaining uncertainties on the difference between requiring the & or not is a c++ issue and not a root problem though, so I’ll study it elsewhere.

Thanks for your help!