soft and program: comp.lang.c++ - 13 new messages in 8 topics

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==============================================================================
TOPIC: 2008 Collection Salina GM M95612 Collection
http://groups.google.com/group/comp.lang.c++/t/6d8637a801b1639d?hl=en
==============================================================================

== 1 of 1 ==
Date: Mon, Dec 29 2008 5:58 pm
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==============================================================================
TOPIC: What does the 'static' in "vector<static FixPt> Data1024;" stand for?
http://groups.google.com/group/comp.lang.c++/t/9c78f686d25b7ce4?hl=en
==============================================================================

== 1 of 2 ==
Date: Mon, Dec 29 2008 6:06 pm
From: Cuthbert

Hi folks,

I am a newcomer to C++ and am reading a example code right now. In
this example code, the author used a lot of "vector<static DataType>
somename;". I know the static vector definition but wonder what's the
different between static vector and "vector<static xxx> xxx"? Can
anybody explain this for me? Thanks a lot.

Cuthbert

== 2 of 2 ==
Date: Mon, Dec 29 2008 7:36 pm
From: alfps

On 30 Des, 03:06, Cuthbert <cuthbert....@gmail.com> wrote:
> I am a newcomer to C++ and am reading a example code right now. In
> this example code, the author used a lot of "vector<static DataType>
> somename;". I know the static vector definition but wonder what's the
> different between static vector and "vector<static xxx> xxx"? Can
> anybody explain this for me? Thanks a lot.

The usage

vector<static T> v;

is not valid standard C++.

If you're sure that the example was /exactly/ like that, then please
tell the group what book you're reading, and locate the nearest
fireplace.

There remains of course the possibility of some compiler-specific
language extension, and/or that you remember the example incorrectly.

Cheers & hth.,

- Alf

==============================================================================
TOPIC: Initializers?
http://groups.google.com/group/comp.lang.c++/t/4eabc57174097969?hl=en
==============================================================================

== 1 of 1 ==
Date: Mon, Dec 29 2008 6:45 pm
From: asm23

Bill wrote:
>
> I think I acquired my understanding from Stroustrup's great book: The C++
> Programming Language. I just tried both versions of the program on VC++
> (version 6.0), and both compiled without a warning. So the bottom line is
> your compiler's requirements--no matter what the language definition says!
> :)
>
> Bill
>
>
Yes.
Thanks Bill for your reply.
It's exciting to discuss on this forum, and I gain my understanding a lot.

==============================================================================
TOPIC: MFC Golmine site updated http://mfcgoldmine.by.ru/index.html
http://groups.google.com/group/comp.lang.c++/t/cd8333fd25b20df2?hl=en
==============================================================================

== 1 of 1 ==
Date: Mon, Dec 29 2008 8:03 pm
From: enter@exit.here (Labyrinth)

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==============================================================================
TOPIC: Gucci Medium Tote Bag Silver 189669 Collection
http://groups.google.com/group/comp.lang.c++/t/15ae58e5a020e03a?hl=en
==============================================================================

== 1 of 1 ==
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TOPIC: Luxury Gucci Queen - Best Gucci Handbags Collection
http://groups.google.com/group/comp.lang.c++/t/9520e0de8b84e8e2?hl=en
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==============================================================================
TOPIC: why vector needs an assignable requirement for object T ?
http://groups.google.com/group/comp.lang.c++/t/6a67019f6c14af86?hl=en
==============================================================================

== 1 of 3 ==
Date: Mon, Dec 29 2008 10:09 pm
From: abir

Hi
I am using an object with a const member like ,
struct Foo{
const int x_;
explicit Foo(int x) : x_(x){}
};
I am happy with default copy ctor ...
but as its member is const, i can't make it assignable.

if i have std::vector<Foo> FV;
FV v; v.push_back(Foo(1));
Now i can't store it in vector. push_back internally calls insert and
which somewhere needs fill.
Why it is required at all,?
I have a handcrafted vector class where push_back does not need
assignment (not even copy ctor, an move op is sufficient)
I don't know where it is required apart from v[i] = Foo(3); or
equivalently *it = Foo(3); kind of statement.
even for insert at the mid i can transfer the mid elements at the
tail and copy /move construct the new elements at the raw memory at
mid.

Can anybody help me here?
Thanks

== 2 of 3 ==
Date: Mon, Dec 29 2008 11:30 pm
From: alfps

On 30 Des, 07:09, abir <abirba...@gmail.com> wrote:
> Hi
> I am using an object with a const member like ,
> struct Foo{
> const int x_;
> explicit Foo(int x) : x_(x){}};
>
> I am happy with default copy ctor ...
> but as its member is const, i can't make it assignable.
>
> if i have std::vector<Foo> FV;
> FV v; v.push_back(Foo(1));
> Now i can't store it in vector. push_back internally calls insert and
> which somewhere needs fill.
> Why it is required at all,?
> I have a handcrafted vector class where push_back does not need
> assignment (not even copy ctor, an move op is sufficient)
> I don't know where it is required apart from v[i] = Foo(3); or
> equivalently *it = Foo(3); kind of statement.
> even for insert at the mid i can transfer the mid elements at the
> tail and copy /move construct the new elements at the raw memory at
> mid.

When you use a non-assignable element type you're violating the
promise that that type gives: if you have a pointer or reference or
iterator to such an element you have a right to expect that it won't
change.

Removing or inserting in a vector can however change the elements
beyond the removal/insertion point.

And there is no separate class for "constant vector".

In the end it's an issue of complexity.

std::vector would have to be more complicated to deal with the issue
of non-assignable element type (but note that the restriction to
assignable is just a specification bug for std::list, corrected in C+
+0x).

Cheers & hth.,

- Alf

== 3 of 3 ==
Date: Tues, Dec 30 2008 12:06 am
From: abir

On Dec 30, 12:30 pm, alfps <alf.p.steinb...@gmail.com> wrote:
> On 30 Des, 07:09, abir <abirba...@gmail.com> wrote:
>
>
>
> > Hi
> > I am using an object with a const member like ,
> > struct Foo{
> > const int x_;
> > explicit Foo(int x) : x_(x){}};
>
> > I am happy with default copy ctor ...
> > but as its member is const, i can't make it assignable.
>
> > if i have std::vector<Foo> FV;
> > FV v; v.push_back(Foo(1));
> > Now i can't store it in vector. push_back internally calls insert and
> > which somewhere needs fill.
> > Why it is required at all,?
> > I have a handcrafted vector class where push_back does not need
> > assignment (not even copy ctor, an move op is sufficient)
> > I don't know where it is required apart from v[i] = Foo(3); or
> > equivalently *it = Foo(3); kind of statement.
> > even for insert at the mid i can transfer the mid elements at the
> > tail and copy /move construct the new elements at the raw memory at
> > mid.
>
> When you use a non-assignable element type you're violating the
> promise that that type gives: if you have a pointer or reference or
> iterator to such an element you have a right to expect that it won't
> change.
I am not sure what promise the type need to give. The only promise i
can think of is that the container can rearrange the elements in any
memory in any order (for vector even the memory order is specified as
for any ith element &v[i+1] == &v[i]+1)
Why the type need to promise that it need to be copy ctor'able apart
from that it can be moved inside the vector's memory (through copy,
through move, through direct construct or whatever other way
possible). I think c++0x removed that requirement, as in latest gcc
build i can write
v.push_back(1,2); for a struct which can take 2 args, and do inplace
construction.
or even a move ctor if i had externally constructed the object.
After that it can only rearrange the elements, or place them in some
new memory etc.
Why i need to give a promise that the vector (or any other container)
can change the state of the object which i requested it to store only?
one can always move an object with const state (or const object) in
another memory.
raw memory
> Removing or inserting in a vector can however change the elements
> beyond the removal/insertion point.
>
> And there is no separate class for "constant vector".
>
> In the end it's an issue of complexity.
>
It may be issue of complexity, but why it is a requirement?
> std::vector would have to be more complicated to deal with the issue
> of non-assignable element type (but note that the restriction to
> assignable is just a specification bug for std::list, corrected in C+
> +0x).
>
> Cheers & hth.,
>
> - Alf

==============================================================================
TOPIC: Is this copy initialization or direct initialization ?
http://groups.google.com/group/comp.lang.c++/t/bbb9bf8b25a8df90?hl=en
==============================================================================

== 1 of 3 ==
Date: Mon, Dec 29 2008 10:27 pm
From: "subramanian100in@yahoo.com, India"

Suppose 'Test' is a class.

Suppose I create an instance 'obj' of the 'Test' class:

Test obj;

Now, suppose I write

Test another = obj;

My question: is this last line called 'direct -initialization' or
'copy-initialization' ?

Kindly clarify.

Thanks
V.Subramanian

== 2 of 3 ==
Date: Mon, Dec 29 2008 10:53 pm
From: amrollahi.saeed@gmail.com

On Dec 30, 9:27 am, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.com> wrote:
> Suppose 'Test' is a class.
>
> Suppose I create an instance 'obj' of the 'Test' class:
>
> Test obj;
>
> Now, suppose I write
>
> Test another = obj;
>
> My question: is this last line called 'direct -initialization' or
> 'copy-initialization' ?
>
> Kindly clarify.
>
> Thanks
> V.Subramanian

AFAIK, there isn't "direct -initialization" terminology in standard C+
+. I think the term "copy-initialization"
isn't standard too. What is clear is, in the following statements:
Test obj;
Test another = obj;
first an object (obj) is created using default constructor, then
another object (another) is created using
copy constructor. The second one is called "initialization", because
another is initialized using obj.

Cheers,
Saeed Amrollahi

== 3 of 3 ==
Date: Mon, Dec 29 2008 11:28 pm
From: blargg.h4g@gishpuppy.com (blargg)

subramanian100in wrote:
> Suppose 'Test' is a class.
>
> Suppose I create an instance 'obj' of the 'Test' class:
>
> Test obj;
>
> Now, suppose I write
>
> Test another = obj;
>
> My question: is this last line called 'direct -initialization' or
> 'copy-initialization' ?

It's copy-initialization, but since obj is the same type as another,
it has the semantics of direct-initialization (that is, it constructs
another using Test's copy constructor, without any temporary). See
section 8.5 paragraphs 12 and 14 (shown below).

ISO/IEC 14882:2003

8.5 Initializers
[...]
11 The form of initialization (using parentheses or =) is generally
insignificant, but does matter when the entity being initialized has a
class type; see below. A parenthesized initializer can be a list of
expressions only when the entity being initialized has a class type.

12 The initialization that occurs in argument passing, function return,
throwing an exception (15.1), handling an exception (15.3), and
brace-enclosed initializer lists (8.5.1) is called copy-initialization
and is equivalent to the form

T x = a;

The initialization that occurs in new expressions (5.3.4), static_cast
expressions (5.2.9), functional notation type conversions (5.2.3), and
base and member initializers (12.6.2) is called direct-initialization
and is equivalent to the form

T x(a);

[...]

14 The semantics of initializers are as follows.
[...]
‹ If the destination type is a (possibly cv-qualified) class type:

‹ If the class is an aggregate (8.5.1), and the initializer is a
brace-enclosed list, see 8.5.1.

‹ If the initialization is direct-initialization, or if it is
copy-initialization where the cv-unqualified version of the source
type is the same class as, or a derived class of, the class of the
destination, constructors are considered. The applicable
constructors are enumerated (13.3.1.3), and the best one is chosen
through overload resolution (13.3). The constructor so selected is
called to initialize the object, with the initializer
expression(s) as its argument(s). If no constructor applies, or
the overload resolution is ambiguous, the initialization is
ill-formed.

‹ Otherwise (i.e., for the remaining copy-initialization cases),
user-defined conversion sequences that can convert from the source
type to the destination type or (when a conversion function is
used) to a derived class thereof are enumerated as described in
13.3.1.4, and the best one is chosen through overload resolution
(13.3). If the conversion cannot be done or is ambiguous, the
initialization is ill-formed. The function selected is called with
the initializer expression as its argument; if the function is a
constructor, the call initializes a temporary of the destination
type. The result of the call (which is the temporary for the
constructor case) is then used to direct-initialize, according to
the rules above, the object that is the destination of the
copy-initialization. In certain cases, an implementation is
permitted to eliminate the copying inherent in this
direct-initialization by constructing the intermediate result
directly into the object being initialized; see 12.2, 12.8.

==============================================================================

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Tuesday, December 30, 2008

comp.lang.c++ - 13 new messages in 8 topics - digest

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