Question

in this code:

  int* a;
  int* b;
  int c;
  int* d;
  a=new int(5);
  b=&(*a);
  c=*a;
  d=&c;
  cout<<"*a = "<<*a<<endl;
  cout<<"a = "<<a<<endl;
  cout<<"b ="<<b<<endl;
  cout<<"d = "<<d<<endl;

I get:

*a = 5
 a = 0x83a2008
 b =0x83a2008
 d = 0xbfbfe540

why is d different from b? Aren’t they both &(*a) ? how can I get the d result with a single line?

Thanks a lot.

Answer 1

a points to a dynamically allocated location, holding the value 5. When you do c = *a;, you’re copying the value 5 from that dynamically allocated location into c. You’re then taking the address of c and assigning it to d (and printing it out).

When you end up with is something like this:

The solid lines indicate a pointer referring to a location. The dashed line indicates movement of data.