• Carlos (14/14) Aug 20 2002 This should raise an exception, right?
• Walter (9/23) Aug 20 2002 No, the invariant only gets called when a public member function gets
• Carlos (4/29) Aug 21 2002 Ok. Thanks.

"Carlos" <carlos8294 msn.com> writes:

This should raise an exception, right?

class A {
  int a;
  invariant {
    assert(a>0);
  }
}

void main()
{
  A b;
  b=new A();
  b.a=-2;
}

Besides, if I declare a as private, it still works. Why?

"Walter" <walter digitalmars.com> writes:

"Carlos" <carlos8294 msn.com> wrote in message
news:aju78e$1r1o$1 digitaldaemon.com...
 This should raise an exception, right?

No, the invariant only gets called when a public member function gets
called, and after the constructor gets called. There is no constructor
declared for A. If you do declare one like:
    this() { }
then the assert will trip.

 class A {
   int a;
   invariant {
     assert(a>0);
   }
 }

 void main()
 {
   A b;
   b=new A();
   b.a=-2;
 }

 Besides, if I declare a as private, it still works. Why?

Because main() is in the same module as class A, so they are 'friends'. If A
is put into a separate module, then you'll get the access violation.

"Carlos" <carlos8294 msn.com> writes:

Ok. Thanks.

"Walter" <walter digitalmars.com> escribió en el mensaje
news:ajv52t$2prk$1 digitaldaemon.com...
 "Carlos" <carlos8294 msn.com> wrote in message
 news:aju78e$1r1o$1 digitaldaemon.com...
 This should raise an exception, right?

 No, the invariant only gets called when a public member function gets
 called, and after the constructor gets called. There is no constructor
 declared for A. If you do declare one like:
     this() { }
 then the assert will trip.

 class A {
   int a;
   invariant {
     assert(a>0);
   }
 }

 void main()
 {
   A b;
   b=new A();
   b.a=-2;
 }

 Besides, if I declare a as private, it still works. Why?

 Because main() is in the same module as class A, so they are 'friends'. If

A
 is put into a separate module, then you'll get the access violation.