How to get the entry of a TH1F?

ROOT
1

Dear rooters:
I have a group of TH1Fs which are generated from data depending on different cut. What I like to do is to get the entry of each histogram. When i used the function GetEntries(), the entry got have no change so it’s not right. When I use the function GetReadEntry(), an error happenned . Below is the section related. Thanks for help.

the erros looks like; Error: Can’t call TH1F::GetReadEntry() in current scope

[code]Double_t cwidth[50];
Int_t outcicount[50],outcocount[50];
Double_t cs;
char hsititle[50],hsotitle[50];

for(int i=0;i<50;i++){
cs=1.5+i/20;
cwidth[i]=cs;
  Double_t z2tlidn=z2tlipk-cs*z2tlisig; Double_t z2tliup=z2tlipk+cs*z2tlisig;
  sprintf(cutexp,"z2ac>=%10.6f && z2ac<=%10.6f",z2tlidn,z2tliup);
TCut staczi=cutexp;
  Double_t eptlidn=eptlipk-cs*eptlisig; Double_t eptliup=eptlipk+cs*eptlisig;
  sprintf(cutexp,"(t2si2e+t2si3e+8.383121*tof2)>=%10.6f && (t2si2e+t2si3e+8.383121*tof2)<=%10.6f",eptlidn,eptliup);
TCut stacEi=cutexp;
  Double_t z2tlodn=z2tlopk-cs*z2tlosig; Double_t z2tloup=z2tlopk+cs*z2tlosig;
  sprintf(cutexp,"z2ac>=%10.6f && z2ac<=%10.6f",z2tlodn,z2tloup);
TCut staczo=cutexp;
  Double_t eptlodn=eptlopk-cs*eptlosig; Double_t eptloup=eptlopk+cs*eptlosig;
  sprintf(cutexp,"(t2si2e+t2si3e+8.383121*tof2)>=%10.6f && (t2si2e+t2si3e+8.383121*tof2)<=%10.6f",eptlodn,eptloup);
TCut stacEo=cutexp;

TCut acidep=stacEi && staczi && cposi;	
TCut acodep=stacEo && staczo && cposi;	

sprintf(hsititle,"C in acpid cut width%4.2f x@target",cs);
sprintf(hsotitle,"C out acpid cut width%4.2f x@target",cs);

acicountx= new TH1F("acicountx",hsititle,25,-25,25);
chain.Draw("target1x>>acicountx",acidep);
outcicount[i]=acicountx->GetReadEntry();

acocountx= new TH1F("acocountx",hsotitle,25,-25,25);
chain.Draw("target1x>>acocountx",acodep);
outcocount[i]=acocountx->GetReadEntry();
}[/code]
2

TH1::GetReadEntry does not exist.
TH1::GetEntries return a Double_t. If you want to store it in an integer,
you must cast it, ie replace

outcocount[i]=acocountx->GetReadEntry(); by

Rene

3

Rene,

Thank you so much. That works.