Hi,
I have always been taught to use an inialization list for initialising data
members of a class. I realize that initialsizing primitives and pointers use
an inialization list is exactly the same as an assignment, but for class
types it has a different effect - it calls the copy constructor.
My question is when to not use an initalisation list for initialising data
members of a class?
Regards
Adi
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tom_usenet wrote:
[SNIP] Yes, when you don't give an initializer, it is still initialized.
"Initialize r" and "Initialization " are different things. Even if you
don't use an initializer, initialization still takes place.
Initialization is the action. An action of giving a determined value to
some object. If this is not done, well, then it is not initialized.
This is a sort of foxymoron. A smart way of telling that if I do not
initialize (give a value) to an object I still initialize (give a value) to
an object. Something is not right here. Later of course we say: well, we
have initialized the value but..., hm, khm, we still cannot use it until we
give a value to it. Something is not right. We have to be brave enough to
say: no initialization happens. The POD is not necessarily itself. That
int might not be int! How could we say: it is initialized? If the standard
says that - it is wrong. We can vote to say the Earth is flat and the fixed
center of the universe it will still not be. :-)
The lvalue to rvalue thing applies to things that have been
initialized too:
int* p = new int;
delete p; //p now indeterminate (but obviously initialized)
Nope. It is invalid, not indetermined.
int* q = p; //error, lvalue-to-rvalue conversion
Totally different case.
And it is still moving... ;-)
--
WW aka Attila
Gary Labowitz wrote:
[SNIP] This i, if it has an automatic storage, is not initialized. Accessing it for anything else then writing (assigment or
initialization)
If it has not been initialized, how can it be initialized later?
I thought that initialization can only occour when a variable
is created. Afterwards it is always assignment.
Would this allow for assigning a value to a const variable during
construction of the object containing the variable?
Nope. My wording was not clear.
(Or does it have
to be initialized in the declaration statement?)
It has to be initialized in an initializer list in the a class, or in case
of a static member in its definition (or inside the class if it is an
integral constant). "Normal" constant variables has to be initialized at
their definition.
Assigning the initial value of a variable (even const, called
"final") is allow in (for example) Java.
OK. In C++ it is called initialization. A const value can only be
initialized and must be initialized. You cannot assign a new value to it
(whatever value means for it).
They appear to keep track of
a variable's state as "uninitiali zed" and define the assigning of a
value during construction as initialization. Does C++ do this?
Nope. The closest thing to this in C++ is the way static variables in a
function scope are initialized. But those are the exception, C++ does nto
keep track of initialized state - in the name of the "cause of all evil",
optimization. :-)
If it
did, then calling a constructor as part of an initialization list is
perfectly valid; and I think it is!
Caling a constructor on the _member_ intiializer list is valid. But not for
the reasons you mention.
--
WW aka Attila
tom_usenet wrote:
[SNIP] If you choose your words carefully it makes more sense: "If we don't
provide an initializer for a POD, the object is initialized to an
indeterminate value."
Initialization == giving an initial value
not giving an initial value (but leave garbage there) != initialization
"initialize " and "provide an initializer for" are best thought of as
different things. initialize is something the compiler (or exe) does
to a variable. It does it whether you provide an initializer or not.
Yes. But the compiler (or the exe) does *nothing* to uninitialized PODs!
--
WW aka Attila
"Ying Yang" <Yi******@hotma il.com> wrote in message news:3f******** @news.iprimus.c om.au... If my class does not extend any class then there would be no base class your
referring to. Maybe you mean a default constructor of a class must always be
present when you use an initialisation list. Can you be more clear?
Some constructor has to be defined to use an initializer list (doesn't necessarily
have to be the default). This only makes sense simce the initialization list is
part of the constructor definition syntax.
He said IF you have a base class with no default constructor. If you derive from
the derived constructors need to provide initializers so that a valid constructor can
be invoked.
On Mon, 15 Sep 2003 16:43:58 +0300, "White Wolf" <wo***@freemail .hu>
wrote: tom_usenet wrote:
[SNIP] Yes, when you don't give an initializer, it is still initialized.
"Initialize r" and "Initialization " are different things. Even if you
don't use an initializer, initialization still takes place.
Initializati on is the action.
Ok
An action of giving a determined value to
some object.
Who says? Not the standard.
struct A
{
int i;
A(int){}
};
A a(1); // initialized? yes. determinate value? no.
If this is not done, well, then it is not initialized.
Your definition of initialized is wrong. e.g. 6.7/2
"Variables with automatic storage duration (3.7.2) are initialized
each time their declaration*sta tement is executed."
It doesn't say that you initialize them. It says that they are
initialized (maybe by you, maybe by the compiler, maybe by the great
goddess of undefined behaviour).
This is a sort of foxymoron. A smart way of telling that if I do not
initialize (give a value) to an object I still initialize (give a value) to
an object.
Why the use of "I"? The compiler can give a value to an object (aka
initialize it). This value might be indeterminate. Here is an example
of just that (from 8.5.1/8):
"An empty initializer*lis t can be used to initialize any aggregate. If
the aggregate is not an empty class, then each member of the aggregate
shall be initialized with a value of the form T() (5.2.3), where T
represents the type of the uninitialized member. "
Here, "initialize d" means "initialize d by the compiler/runtime",
whereas the "uninitiali zed" means "not directly initialized by an
initializer". The lvalue to rvalue thing applies to things that have been
initialized too:
int* p = new int;
delete p; //p now indeterminate (but obviously initialized)
Nope. It is invalid, not indetermined.
Ahh, yes, although it might be useful to allow it to fall under 4.1 by
changing the language in both places to "inderminat e value" or
similar.
The standard seems to be a bit sloppy about its use of the word
"initialize " and this could probably do with clearing up...
Tom
On Mon, 15 Sep 2003 16:51:15 +0300, "White Wolf" <wo***@freemail .hu>
wrote: tom_usenet wrote:
[SNIP] If you choose your words carefully it makes more sense: "If we don't
provide an initializer for a POD, the object is initialized to an
indeterminate value."
Initializati on == giving an initial value
Indeed, however this doesn't have to involve an initializer.
not giving an initial value (but leave garbage there) != initialization
Nope, the compiler gives an initial value to PODs if you don't bother.
This is also initialization.
"initialize " and "provide an initializer for" are best thought of as
different things. initialize is something the compiler (or exe) does
to a variable. It does it whether you provide an initializer or not.
Yes. But the compiler (or the exe) does *nothing* to uninitialized PODs!
Are you saying that the bytes making up the value representation of a
POD have to be changed for initialization to have occurred? I don't
think the standard backs you on this!
Tom
"foo" <ma*******@axte r.com> wrote in message
news:c1******** *************** **@posting.goog le.com... "jeffc" <no****@nowhere .com> wrote in message
news:<3f******* *@news1.prserv. net>... "Erik" <no@spam.com> wrote in message news:bj******** **@news.lth.se. .. > The basic rule of thumb is that anything that can go in the
initialize list > should go there. Obviously, you can't put something like this in
there: > if (a)
> i = 2;
> else
> i = 4;
Many, if not most, of those cases can be covered by the ?: operator:
MyClass(int a) : i(a ? 2 : 4) {}
Yeah, maybe simple "if" cases like that, but you can't put logic or
function calls (other than base class constructors) in the initializer list. You
added an expression, not a statement.
There are plenty of things you can put in the initializer list.
You can put functions, additions, conditon logic, and more.
And there are plenty of things you can't. You can put expressions in there,
but you can't put statements in there. You can't call functions (alone),
you can't use for or while loops, you can't... well, you already know what
you can't do.
"Ron Natalie" <ro*@sensor.com > wrote in message
news:3f******** *************** @news.newshosti ng.com...
"jeffc" <no****@nowhere .com> wrote in message
news:3f******** @news1.prserv.n et...
Yep - and sometimes it makes sense to create a function that does more
complex things so that everything is in the initializer list.
And how exactly do you propose calling such a function from the
initializer list?
Same way you call any other function.
Yeah, right.
"Ron Natalie" <ro*@sensor.com > wrote in message
news:3f******** *************** @news.newshosti ng.com...
"jeffc" <no****@nowhere .com> wrote in message
news:3f******** @news1.prserv.n et... Many, if not most, of those cases can be covered by the ?: operator:
MyClass(int a) : i(a ? 2 : 4) {}
Yeah, maybe simple "if" cases like that, but you can't put logic or
function calls (other than base class constructors) in the initializer list. You
added an expression, not a statement.
Who says you can't have function calls?
You can only have function calls as part of expressions, not statements.
You can't simply call a function such as initialize() as Gianni suggested
(as you well know.)
On Mon, 15 Sep 2003 11:29:54 -0400, "jeffc" <no****@nowhere .com>
wrote:
"Ron Natalie" <ro*@sensor.com > wrote in message
news:3f******* *************** *@news.newshost ing.com...
"jeffc" <no****@nowhere .com> wrote in message
news:3f******* *@news1.prserv. net... > > Many, if not most, of those cases can be covered by the ?: operator:
> > MyClass(int a) : i(a ? 2 : 4) {}
>
> Yeah, maybe simple "if" cases like that, but you can't put logic orfunction > calls (other than base class constructors) in the initializer list. You
> added an expression, not a statement.
Who says you can't have function calls?
You can only have function calls as part of expressions, not statements.
You can't simply call a function such as initialize() as Gianni suggested
(as you well know.)
I think you completely misunderstood him:
struct A
{
int a;
static int doCalc(int val);
A(int val): a(doCalc(val))
{
}
};
That was exactly the kind of thing that Gianni was suggesting - if you
can't put the logic inline, make a function for it. If your member is
const or a reference, you have no choice but to do so.
Tom This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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