Simple constructor question

[Christopher Benson-Manica]

(if this is a FAQ, I apologize - I didn't see it)

I have the following class declaration:

class TMSViewDayList : public TMSViewDayListB ase {
public:
uint start;
uint end;
TMSViewDayList( ) : start(0),end(0) {}
};

TMSViewDayListB ase is a template class that acts something like a vector. My
question is, is TMSViewDayListB ase's default constructor implicitly called
here, or do I need to explicitly call it like

TMSViewDayList( ) : start(0),end(0) ,TMSViewDayList Base() {}

?

--
Christopher Benson-Manica | Upon the wheel thy fate doth turn,
ataru(at)cybers pace.org | upon the rack thy lesson learn.

[Gianni Mariani]

Re: Simple constructor question

Christopher Benson-Manica wrote:[color=blue]
> (if this is a FAQ, I apologize - I didn't see it)
>
> I have the following class declaration:
>
> class TMSViewDayList : public TMSViewDayListB ase {
> public:
> uint start;
> uint end;
> TMSViewDayList( ) : start(0),end(0) {}
> };
>
> TMSViewDayListB ase is a template class that acts something like a vector. My
> question is, is TMSViewDayListB ase's default constructor implicitly called
> here, or do I need to explicitly call it like
>
> TMSViewDayList( ) : start(0),end(0) ,TMSViewDayList Base() {}[/color]

It's implicit.

Easy to test with a few lines of example code.

The order of evaluation is interesting as well.

[jeffc]

Re: Simple constructor question


"Christophe r Benson-Manica" <ataru@nospam.c yberspace.org> wrote in message
news:bm12sq$asq $1@chessie.cirr .com...[color=blue]
> (if this is a FAQ, I apologize - I didn't see it)
>
> I have the following class declaration:
>
> class TMSViewDayList : public TMSViewDayListB ase {
> public:
> uint start;
> uint end;
> TMSViewDayList( ) : start(0),end(0) {}
> };
>
> TMSViewDayListB ase is a template class that acts something like a vector.[/color]
My[color=blue]
> question is, is TMSViewDayListB ase's default constructor implicitly called[/color]

Yes.

[David Rubin]

Re: Simple constructor question

Gianni Mariani wrote:

[snip][color=blue]
> The order of evaluation is interesting as well.[/color]

This is interesting. For example:

class B
{
protected:
int i;
public:
B() : i(10) {}
};

class A : public B
{
int j;
public:
A() : i(20), j(30) {}
};

int
main()
{
A a;
return 0;
}

; g++ derive.cc
derive.cc: In constructor `A::A()':
derive.cc:14: error: class `A' does not have any field named `i'

What is going on here? Assigning i inside the body of A() works, but not
in the initializer list. Invoking B() in the initialization list before
i(20) does not work either.

/david

--
Andre, a simple peasant, had only one thing on his mind as he crept
along the East wall: 'Andre, creep... Andre, creep... Andre, creep.'
-- unknown

[jeffc]

Re: Simple constructor question


"David Rubin" <bogus_address@ nomail.com> wrote in message
news:3F845A96.C 44F3250@nomail. com...[color=blue]
> Gianni Mariani wrote:
>
> [snip][color=green]
> > The order of evaluation is interesting as well.[/color]
>
> This is interesting. For example:
>
> class B
> {
> protected:
> int i;
> public:
> B() : i(10) {}
> };
>
> class A : public B
> {
> int j;
> public:
> A() : i(20), j(30) {}
> };
>
> int
> main()
> {
> A a;
> return 0;
> }
>
> ; g++ derive.cc
> derive.cc: In constructor `A::A()':
> derive.cc:14: error: class `A' does not have any field named `i'
>
> What is going on here? Assigning i inside the body of A() works, but not
> in the initializer list. Invoking B() in the initialization list before
> i(20) does not work either.[/color]

i doesn't belong in the initializer list for A because it's not a member of
A. You can assign to i inside the body of A, yes, but you said "not in the
initializer list". There's a difference between initializing and assigning
that's important in this case. It's called the "initialize r list" instead
of the "assignment list" because it's initialization, not assignment. If
you assign to i in the body of A, it's already been initialized by B.

[David Rubin]

Re: Simple constructor question

jeffc wrote:

[snip][color=blue][color=green]
> > What is going on here? Assigning i inside the body of A() works, but not
> > in the initializer list. Invoking B() in the initialization list before
> > i(20) does not work either.[/color]
>
> i doesn't belong in the initializer list for A because it's not a member of
> A. You can assign to i inside the body of A, yes, but you said "not in the
> initializer list". There's a difference between initializing and assigning
> that's important in this case. It's called the "initialize r list" instead
> of the "assignment list" because it's initialization, not assignment. If
> you assign to i in the body of A, it's already been initialized by B.[/color]

The confusion is that A inherits member i from B. Therefore, why is i
not a member of A when A is being constructed? It would follow that if i
is a member of A, i can be initialized in A(), but this is clearly not
the case.

/david

--
Andre, a simple peasant, had only one thing on his mind as he crept
along the East wall: 'Andre, creep... Andre, creep... Andre, creep.'
-- unknown

[Ron Natalie]

Re: Simple constructor question


"David Rubin" <bogus_address@ nomail.com> wrote in message news:3F84740E.1 024DEAF@nomail. com...
[color=blue]
> The confusion is that A inherits member i from B. Therefore, why is i
> not a member of A when A is being constructed? It would follow that if i
> is a member of A, i can be initialized in A(), but this is clearly not
> the case.[/color]

Inheritance is not the issue. The initializer list is restricted to DIRECT bases,
VIRUTAL bases, and members of the class being initialized. Members of
base classes (or non-direct base classes) are not eligible.

There's some ambiguity here. Which value should B::i be initialized with 10 or 20?

[Julián Albo]

Re: Simple constructor question

David Rubin escribió:
[color=blue]
> The confusion is that A inherits member i from B. Therefore, why is i
> not a member of A when A is being constructed? It would follow that if i
> is a member of A, i can be initialized in A(), but this is clearly not
> the case.[/color]

It is a member of B, then it gets initialized during the initialization
of B. The if initilize it in the constructor of A, it will be
initialized two times, and that is not desired in C++.

Regards.

[jeffc]

Re: Simple constructor question


"David Rubin" <bogus_address@ nomail.com> wrote in message
news:3F84740E.1 024DEAF@nomail. com...[color=blue][color=green]
> >
> > i doesn't belong in the initializer list for A because it's not a member[/color][/color]
of[color=blue][color=green]
> > A. You can assign to i inside the body of A, yes, but you said "not in[/color][/color]
the[color=blue][color=green]
> > initializer list". There's a difference between initializing and[/color][/color]
assigning[color=blue][color=green]
> > that's important in this case. It's called the "initialize r list"[/color][/color]
instead[color=blue][color=green]
> > of the "assignment list" because it's initialization, not assignment.[/color][/color]
If[color=blue][color=green]
> > you assign to i in the body of A, it's already been initialized by B.[/color]
>
> The confusion is that A inherits member i from B. Therefore, why is i
> not a member of A when A is being constructed? It would follow that if i
> is a member of A, i can be initialized in A(), but this is clearly not
> the case.[/color]

But i is *never* a member of A, not when A is being constructed, and not
after it's been constructed. A does have access rights to i, but i is a
member of B, not of A. I don't think it's accurate to say that A inherits
member i from B (someone correct me if I'm wrong). But even if it were OK
to use the terminology that A inherits i from B, you still would not say
that i is a member of A.

[jeffc]

Re: Simple constructor question


"Ron Natalie" <ron@sensor.com > wrote in message
news:3f8478ae$0 $36946$9a6e19ea @news.newshosti ng.com...[color=blue]
>
> "David Rubin" <bogus_address@ nomail.com> wrote in message[/color]
news:3F84740E.1 024DEAF@nomail. com...[color=blue]
>[color=green]
> > The confusion is that A inherits member i from B. Therefore, why is i
> > not a member of A when A is being constructed? It would follow that if i
> > is a member of A, i can be initialized in A(), but this is clearly not
> > the case.[/color]
>
> Inheritance is not the issue. The initializer list is restricted to[/color]
DIRECT bases,[color=blue]
> VIRUTAL bases, and members of the class being initialized.[/color]

But inheritance is the issue from his point of view, because he believes i
IS a member of the class being initialized (precisely because A inherits
from B.)

[Ron Natalie]

Re: Simple constructor question


"Julián Albo" <JULIANALBO@ter ra.es> wrote in message news:3F847A2E.3 2CBC0A8@terra.e s...
[color=blue]
> It is a member of B, then it gets initialized during the initialization
> of B. The if initilize it in the constructor of A, it will be
> initialized two times, and that is not desired in C++.[/color]

Not only not desired, it's not even possible.

[Julián Albo]

Re: Simple constructor question

Ron Natalie escribió:
[color=blue][color=green]
> > It is a member of B, then it gets initialized during the initialization
> > of B. The if initilize it in the constructor of A, it will be
> > initialized two times, and that is not desired in C++.[/color]
> Not only not desired, it's not even possible.[/color]

Yeah, and it's not possible because the designers of the language found
did not desire it. I was trying to explain why is not allowed.

Regards.