On 10/18/13, Jonathan M Davis wrote:
> If you want to call a member function from an invariant, it
> should be static, or it should be a free function.
I can also be private, the diagnostic is pretty clear about that. :)
On Friday, 18 October 2013 at 09:08:53 UTC, Jonathan M Davis
wrote:
On Friday, October 18, 2013 11:04:25 bearophile wrote:
simendsjo:
> See topic. Why is this not allowed? The function in question
> is
> not virtual.
>
> struct S {
>
> void someFunction() const {}
> const invariant(
On Friday, October 18, 2013 11:04:25 bearophile wrote:
> simendsjo:
> > See topic. Why is this not allowed? The function in question is
> > not virtual.
> >
> > struct S {
> >
> > void someFunction() const {}
> > const invariant() { someFunction(); }
> >
> > }
> > void main() {
> >
> >
simendsjo:
See topic. Why is this not allowed? The function in question is
not virtual.
struct S {
void someFunction() const {}
const invariant() { someFunction(); }
}
void main() {
S s;
}
It being not virtual is not important. In what cases is
invariant() called, simendsjo? I s
See topic. Why is this not allowed? The function in question is
not virtual.
struct S {
void someFunction() const {}
const invariant() { someFunction(); }
}
void main() {
S s;
}
