What happens at the compiler level when I dereference a pointer to pass it by reference?

I'm interested in what the compiler does when I dereference a pointer explicitly to pass it by reference:

void foo(A& arg)
{
   arg.member = 7;
}
void goo()
{
   A* ob = new A();
   foo(*ob);
   printf(ob->member); // should show 7
}

The reason I'm interested is because I believe bringing the dereference out of the function call would create very different behavour:

void foo(A& arg)
{
   arg.member = 7;
}
void goo()
{
   A* ob = new A();
   A ob_dereferenced = *ob;
   foo(ob_dereferenced);
   printf(ob->member); // should show whatever A initialises member to
}

Your comparison is not a fair one because here in the second version:

> void goo()
> {
> A* ob = new A();
> A ob_dereferenced = *ob; // <---- here !!!
> foo(ob_dereferenced);
> printf(ob->member); // should show whatever A initialises member to
> }

You are making a copy of the A instance. Depending on what A is, creating a second instance can have observable side effects.

However, lets put such observable side effects aside, and use

 struct A { int member = 0; };

then there is no observable difference between:

 void moo(A* a) { 
      foo(*a);    
 }

and using a local reference:

 void moo(A* a) { 
      A& b = *a;  
      foo(b);
 }

Those two moo do the same. There is no reason to expect different output from the compiler when optimizations are turned on.

PS: Your both versions leak memory. If you need a pointer for an example you do not need new. For example A a; A* aptr = &a; makes up for a totally fine pointer to A and you dont need to worry about leaks. Above I avoided the issue, by simply assuming the pointer comes from somewhere.