Hello,
I'm learning C++ for a couple of days and play a bit with the
algorithms provided in the STL. One thing I don't understand is the
fact that classes inherited of functors have to be defined using
structs.
The code
template<class type> struct Print : public unary_function< type, void> {
void operator()(type & x) {
cout << x << endl;
}
};
in conjunction with
for_each(childr en.begin(), children.end(), Print<Node<type >*>());
does only work with struct but not with class, i.e.
template<class type> class Print : public unary_function< type, void> {
leads to a compiler error. I thought that classes and structs are more
or less equal, esp. since unary_function is a class itself, so why
can't I just inherit of it?
Thanks for explanations,
Michael 8 4531 mi************* @gmail.com wrote: Hello,
I'm learning C++ for a couple of days and play a bit with the algorithms provided in the STL. One thing I don't understand is the fact that classes inherited of functors have to be defined using structs.
They do not. The code
template<class type> struct Print : public unary_function< type, void> { void operator()(type & x) { cout << x << endl; } };
in conjunction with
for_each(childr en.begin(), children.end(), Print<Node<type >*>());
does only work with struct but not with class, i.e.
template<class type> class Print : public unary_function< type, void> {
leads to a compiler error. I thought that classes and structs are more or less equal, esp. since unary_function is a class itself, so why can't I just inherit of it?
You can. struct and class is equivalent except for accessibility. The
problem most likely is that you forgot to make your operator() public. Thanks for explanations, Michael
/Peter mi************* @gmail.com wrote: Hello,
I'm learning C++ for a couple of days and play a bit with the algorithms provided in the STL. One thing I don't understand is the fact that classes inherited of functors have to be defined using structs.
The code
template<class type> struct Print : public unary_function< type, void> { void operator()(type & x) { cout << x << endl; } };
in conjunction with
for_each(childr en.begin(), children.end(), Print<Node<type >*>());
does only work with struct but not with class, i.e.
template<class type> class Print : public unary_function< type, void> {
Did you make the operator() public?
--
Ian Collins.
Hello,
forgot to make operator() public, as both of you mentioned.
Thanks for the help!,
Michael
<mi************ *@gmail.com> skrev i meddelandet
news:11******** **************@ m73g2000cwd.goo glegroups.com.. . Hello,
forgot to make operator() public, as both of you mentioned.
Which is why many prefer to use a struct in the first place. If
everything is supposed to be public, that is the easiest.
Bo Persson
Bo Persson wrote: Hello,
forgot to make operator() public, as both of you mentioned.
Which is why many prefer to use a struct in the first place. If everything is supposed to be public, that is the easiest.
However, in classes, I always put the public members first, too, so every
class begins with something like:
class Whatever
{
public:
It seems more logical to me to put the members in order of the number of
users, and so I get public first, then protected and at the end private.
Then, when defining a struct, I also get that urge to write the 'public' on
inheritance and after the '{'.
The following are equivalent:
struct Base : Derived
{
int i;
};
class Base : public Derived /* Note the inheritance also */
{
public:
int i;
};
As are the following:
struct Base : private Derived
{
private:
int i;
};
class Base : Derived
{
int i;
};
--
Frederick Gotham
In message <wj************ *******@news.in digo.ie>, Frederick Gotham
<fg*******@SPAM .com> writes The following are equivalent:
struct Base : Derived { int i; };
class Base : public Derived /* Note the inheritance also */ { public:
int i;
};
So your derived struct/class is called Base, and the base one is called
Derived. Hmmm.
--
Richard Herring
Richard Herring posted: In message <wj************ *******@news.in digo.ie>, Frederick Gotham <fg*******@SPA M.com> writes The following are equivalent:
struct Base : Derived { int i; };
class Base : public Derived /* Note the inheritance also */ { public:
int i;
};
So your derived struct/class is called Base, and the base one is called Derived. Hmmm.
At it's not a logic error in the computer sense of the term ; )
--
Frederick Gotham This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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