c++ - Am I actually storing the object in the cpu register here? -

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this question has answer here:

here in program using volatile register object, storing object register here?
why getting address of object 1? please share thought on this.

#include <iostream> using namespace std;   class a{     int i,j,k[999];     long double  arr[9999999];     public:         a(){             i=77; j=89;             cout<<"\nctor\n";         }         void disp()volatile {             cout<<"\ni = "<<i<<" j = "<<j<<"\n";         //  delete this;         }         ~a(){             cout<<"\ndtor\n";         } };   int main(){     register volatile *ao = new a;     cout<<"address of = "<<ao; //out puts "1" me; (my processor core i3 330m).     ao->disp();      delete ao; }

my thought: here might trying deny optimization of compiler pointer object ao using volatile keyword, right? reason why got '1'! c++ standard says, if specify object or variable register internally consider auto. detaild info out put 1 see link


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