c++ - Why the dangling pointer is giving the size of the previously pointed variable? -

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• can local variable's memory accessed outside scope? 20 answers

class b { public:     b():a(0), b(0) { }     b(int x):a(x), b(0) { } private:     int a;     int b; };  class { public:     a(b* ptr):pb(ptr) { }     void modifypb()      {          delete pb;          pb = null;      }     void printbsize()      {          if( pb != null )             cout<<"pb pointing obj size:"<<sizeof(*pb)<<endl;          else             cout<<"pb pointing obj size:"<<sizeof(*pb)<<endl;      }  private:     b *pb; };   void main() {     b *bobj = new b(10);     cout<<"size of bobj:"<<sizeof(*bobj)<<endl;      aobj(bobj);     cout<<"size of aobj:"<<sizeof(aobj)<<endl;      cout<<"before de-allocating: ";     aobj.printbsize();     aobj.modifypb();     cout<<"after de-allocating: ";     aobj.printbsize(); } 

output:

size of bobj: 8 size of aobj: 4 before de-allocating: pb pointing obj size: 8 after de-allocating: pb pointing obj size: 8 

why size of *pb 8, after de-allocation ?

why size of *pb 8, after de-allocation ?

sizeof(*pb) evaluated @ compile time based on type of *pb. value not depend on value of pb @ run time.

you printing sizeof(b) in both branches of if statement, 8 on platform.