Problem with branch class

ROOT
1

Dear Rooters

Maybe this is a C++ question:
I have a class MyClass and its inherited class MySubClass which are both used for:
tree->Branch(“MyBranch”, “MySubClass”, &mysubclass);
tree->Branch(“MyBranch”, “MyClass”, &myclass);

Now I have a method Export() where I export the content of the trees to a file:

void Export()
{
   TTree *tree;
   MySubClass *myclass;

   tree->SetBranchAddress("MyBranch", &myclass);

   for (Int_t k=0; k<n; k++) {
      tree->GetEntry(i);

      output << sep << myclass->GetValue();
      if (hasSubValue) output << sep << myclass->GetSubValue();
   }
   output << endl;
}

Although this works in both cases, i.e the tree content is always exported to
a file, sorrowly, I get the following error message when I apply Export() to
a tree containing MyClass and not MySubClass:
Error in TTree::SetBranchAddress: The pointer type give (MySubClass) does not correspond to the class needed (MyClass) by the branch: MyBranch

The following workarounds do not work:

a, if (!hasSubValue) myclass = (MyClass*)myclass;
This causes a compiler error:
error: invalid conversion from ‘MyClass*’ to ‘MySubClass*’

b, if (hasSubValue) myclass = (MySubClass*)myclass;
This causes a compiler error:
error: ‘class MyClass’ has no member named ‘GetSubValue’

Does someone have an idea how I could solve this problem?

Thank you in advance
Best regards
Christian

2

[quote]Maybe this is a C++ question: [/quote]Indeed.

[quote]Error in TTree::SetBranchAddress: The pointer type give (MySubClass) does not correspond to the class needed (MyClass) by the branch: MyBranch
[/quote]is a correct message you can not use a MyClass as a MySubClass. But you can use a MySubClass as a MyClass. Hence you should be using:

[code]void Export()
{
TTree *tree;
MyClass *myclass;

tree->SetBranchAddress(“MyBranch”, &myclass);

for (Int_t k=0; k<n; k++) {
tree->GetEntry(i);

  output << sep << myclass->GetValue();

}
output << endl;
} [/code]And to deal with the different printout you should add a virtual function to MyClass and MySubClass which does the correct printing and use as:

[code]void Export()
{
TTree *tree;
MyClass *myclass;

tree->SetBranchAddress(“MyBranch”, &myclass);

for (Int_t k=0; k<n; k++) {
tree->GetEntry(i);
myclass.Output(output);
}
output << endl;
} [/code]

Cheers,
Philippe.

3

Dear Philippe

Thank you for your fast reply.
This is indeed an elegant solution and I will try to implement it.
However, I am not sure that it will be possible, since my real problem is more complex.
Another solution would be to have a virtual function:
myclass->GetValue(“mysubvalue”) which will always return the correct value.

Is there another solution which does not involve adding a virtual function to MyClass?

Best regards
Christian

4

[quote]Another solution would be to have a virtual function:
myclass->GetValue(“mysubvalue”) which will always return the correct value.
[/quote]I am a little confused … Why isn’t myclass->GetValue() sufficient (you would write MyClass::GetValue and MySubClass::GetValue)?

[quote]However, I am not sure that it will be possible, since my real problem is more complex.
…
Is there another solution which does not involve adding a virtual function to MyClass?
[/quote]It depends on what your real problem is! But unless you are trying to retrofit code you can not modify, adding a virtual function is the right solution (for what you described so far)!

Cheers,
Philippe.

5

The problem is that I am exporting the entries of few hundred trees as pivot table,
so I loop first over each tree and then over each of about hundredthousand
entries. For this reason the function should be fast. Probably I will do something like:

Double_t MySubClass::GetValue(Int_t id)
{return (id == 2) ? fVar2 : (id == 3) ? fVar3 : fVar1;}

Thank you
Christian

6

Hi Christian,

I still have too little information to know if this the best solution. Could you be more specific on your class layout?

Cheers,
Philippe.

7

Dear Philippe

It seems to be sufficient to stick to the example since the solution turned out to be simple:

void Export()
{
   TTree *tree;
   MyClass *myclass;

   tree->SetBranchAddress("MyBranch", &myclass);

   for (Int_t k=0; k<n; k++) {
      tree->GetEntry(i);

      output << sep << myclass->GetValue();
      if (hasSubValue) output << sep << (MySubClass*)myclass->GetSubValue();
   }
   output << endl;
}

Sorrowly, all the other solutions do not seem to work.

Thank you for your help and sorry for the confusion.

Best regards
Christian