Is the behavior of return x++; defined?

It is defined.

It returns the value of x before incrementation. If x is a local(non-static) variable this post incrementation has no effect since local variables of a function cease to exist once the function returns. But if x is a local static variable, global variable or an instance variable( as in your case), its value will be incremented after return.

Yes.

In postincrement (x++) the value of x is evaluated (returned in your case) before 1 is added.

In preincrement (++x) the value of x is evaluated after 1 is added.

Edit: You can compare the definition of pre and post increment in the links.

Yes, it's equivalent to:

int bar()
{
  int temp = x;
  ++x;
  return temp;
}

Yes it is ... it will return the x's value before incrementing it and after that the value of x will be + 1 ... if it matters.