soft and program: comp.lang.c++ - 25 new messages in 17 topics

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* please help me in this area(returning class using recursion)...... - 6
messages, 2 authors
http://groups.google.com/group/comp.lang.c++/t/738a7434e286543a?hl=en
* Run-time-checking whether a pure virtual function has been implemented - 2
messages, 2 authors
http://groups.google.com/group/comp.lang.c++/t/71c3c8c3041ae524?hl=en
* canceling noncopyable feature - 2 messages, 2 authors
http://groups.google.com/group/comp.lang.c++/t/5add783081f15e0e?hl=en
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==============================================================================
TOPIC: please help me in this area(returning class using recursion)......
http://groups.google.com/group/comp.lang.c++/t/738a7434e286543a?hl=en
==============================================================================

== 1 of 6 ==
Date: Fri, Aug 28 2009 4:55 am
From: Francesco

On 28 Ago, 13:37, Juha Nieminen <nos...@thanks.invalid> wrote:
> Francesco wrote:
> > s=s*p/(p=s);
> > ...and being that the / operator associates left to right, i.e it
> > evaluates "s*p" first and "(p=s)" last
>
> I would be surprised if the standard required for the left part to be
> evaluated before the right part. As long as the result of the expression
> is correct, it doesn't matter in which order the elements are evaluated.

Pulling apart that I mistaken the meaning of "association" - because
it involves grouping and not evaluation order - later on I thought
that the two involved operands _ensured_ the evaluation order. Looking
at the operators precedence, I would have said that "(p=s)" _had_ to
be evaluated first, just like James said. The problem is that my GCC
evaluates "s*p" first. And it seems to do so systematically. I'm about
to test it further.

Well, after all I'm not the OP and all of this is just a chance to
learn new things.

Best regards,
Francesco

== 2 of 6 ==
Date: Fri, Aug 28 2009 5:02 am
From: James Kanze

On Aug 28, 11:47 am, Francesco <entul...@gmail.com> wrote:
> On 28 Ago, 10:32, James Kanze <james.ka...@gmail.com> wrote:
> > On Aug 27, 9:17 pm, Juha Nieminen <nos...@thanks.invalid> wrote:

> > > manohar wrote:
> > > >http://pastebin.com/m4338968e
> > > This:
> > > small=small*ptr[i]/(ptr[i]=small);
> > > is not only needlessly obfuscated, but probably also
> > > undefined behavior. (And even if it wasn't, I would never
> > > recommend writing such code. Even after watching it for
> > > minutes I can't tell what is it that that line is trying
> > > to do.)

> > It's definitely undefined behavior. And like you, I can't
> > figure out what it's supposed to be---if you separate it
> > into two expressions:
> > ptr[ i ] = small ;
> > small = small * ptr[ i ] / ptr[ i ] ;
> > it doesn't make much sense, either (especially the last one).

> Hi, sorry again for dropping in out of blue, I just want to
> understand this UB thing.

In a way, it's pure standardese. The standard uses it whenever,
for whatever reason, it doesn't want to constrain an
implementation. In the concrete case above, for example, it's
highly unlikely that you risk reformatting the hard disk, even
though the standard allows it; on most (maybe all) machines, the
standard could just as easily said that the results of the
expression are unspecified.

> First of all, this line:
> small=small*ptr[i]/(ptr[i]=small);
> seems just a weird way to swap the variables.

But it doesn't swap anything.

> Simplifying the names...
> s=s*p/(p=s);
> ...and being that the / operator associates left to right,

Don't confuse association with order of evaluation. An
expression like a + b + c is (a + b) + c, and not a + (b + c),
but the compiler is still free to evaluate a, b and c in any
order, and to cause any side effects of a, b and c to occur at
any time it wishes before the next sequence point.

> i.e it evaluates "s*p" first and "(p=s)" last,

No.

> it would give something like:
> sp = s*p;
> s = sp/(p=s);

> Or in other (less clear) "words"...
> t = s;
> // s=s*p/(p=s); // since "(p=s) == s" this becomes...
> // s=s*p/s; // since "s*p/s == p" this becomes...
> s = p;
> p = t;

> Now the question: does the associativity of operators enter in
> account when considering the UB issue "evaluated twice in the
> same expression"
> - or whatever it is called?

Associativity controls the results of the expression. Consider:

int i = 1 ;
int j = 3 ;
int k = 100 ;

int x = i * k / j ;

Associativity guarantees that x will be equal to 33, and not 0.
If you replace i, j, and k with functions that return the value,
but have side effects (e.g. int i() { std::cout << 'i' ; return
1 ; }), however, there's absolutely no guarantee with regards to
the order these functions will be called. All associativity
guarantees is that the multiplication will take place before the
division (and thus, will not use the results of the division,
but rather only the results of k).

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

== 3 of 6 ==
Date: Fri, Aug 28 2009 5:07 am
From: James Kanze

On Aug 28, 1:35 pm, Juha Nieminen <nos...@thanks.invalid> wrote:
> Francesco wrote:
> > By the way, I recall having read something about a trick
> > used in Assembly, where you swap two variables without using
> > a temporary... is it related to the above expression or to
> > its compilation? I guess they are completely different
> > things.

> It involves doing three bitwise xor operations. It's
> completely counter-productive because the xors will usually be
> slower, or at most as fast, as the classical way of swapping.

And it doesn't always work:

template< typename T >
void
swap( T& a, T& b )
{
a ^= b ;
b ^= a ;
a ^= b ;
}

Given this:

char* p1 ;
char* p2 ;
swap( p1, p2 ) ;

doesn't compile. And:

int a = 42 ;
swap( a, a ) ;

compiles, but gives the wrong answer.

== 4 of 6 ==
Date: Fri, Aug 28 2009 6:12 am
From: Francesco

On 28 Ago, 14:02, James Kanze <james.ka...@gmail.com> wrote:
> On Aug 28, 11:47 am, Francesco <entul...@gmail.com> wrote:
>
>
>
> > On 28 Ago, 10:32, James Kanze <james.ka...@gmail.com> wrote:
> > > On Aug 27, 9:17 pm, Juha Nieminen <nos...@thanks.invalid> wrote:
> > > > manohar wrote:
> > > > >http://pastebin.com/m4338968e
> > > > This:
> > > > small=small*ptr[i]/(ptr[i]=small);
> > > > is not only needlessly obfuscated, but probably also
> > > > undefined behavior. (And even if it wasn't, I would never
> > > > recommend writing such code. Even after watching it for
> > > > minutes I can't tell what is it that that line is trying
> > > > to do.)
> > > It's definitely undefined behavior. And like you, I can't
> > > figure out what it's supposed to be---if you separate it
> > > into two expressions:
> > > ptr[ i ] = small ;
> > > small = small * ptr[ i ] / ptr[ i ] ;
> > > it doesn't make much sense, either (especially the last one).
> > Hi, sorry again for dropping in out of blue, I just want to
> > understand this UB thing.
>
> In a way, it's pure standardese. The standard uses it whenever,
> for whatever reason, it doesn't want to constrain an
> implementation. In the concrete case above, for example, it's
> highly unlikely that you risk reformatting the hard disk, even
> though the standard allows it; on most (maybe all) machines, the
> standard could just as easily said that the results of the
> expression are unspecified.

Fine, the general UB concept is clear to me, I expressed myself badly.
I meant that I'd like to understand this particular case leading to
UB.

>
> > First of all, this line:
> > small=small*ptr[i]/(ptr[i]=small);
> > seems just a weird way to swap the variables.
>
> But it doesn't swap anything.

Actually, it does swap the variables on my GCC version, see below.

> > Simplifying the names...
> > s=s*p/(p=s);
> > ...and being that the / operator associates left to right,
>
> Don't confuse association with order of evaluation. An
> expression like a + b + c is (a + b) + c, and not a + (b + c),
> but the compiler is still free to evaluate a, b and c in any
> order, and to cause any side effects of a, b and c to occur at
> any time it wishes before the next sequence point.

Yes, I confused them - but I fixed my mistake shortly after. Actually,
seems that the * and / operators enforce some kind of sequence point
on my GCC. See below.

> > i.e it evaluates "s*p" first and "(p=s)" last,
>
> No.

As I said above, it does so, on my GCC.

[ associativity details snipped, thanks for the clarification ]

Given the standard specifications, the fact that the following code...

-------
#include <iostream>
using namespace std;
int A(){cout << "A"; return 1;}
int B(){cout << "B"; return 1;}
int C(){cout << "C"; return 1;}
int main()
{
A() * B() / C();
cout << endl;
A() * (B()) / ((C()));
cout << endl;
int s = 1;
int p = 2;
cout << "s==" << s << " p==" << p << endl;
s=s*p/(p=s);
cout << "s==" << s << " p==" << p << endl;
return 0;
}
-------

...produces this output...

-------
ABC
ABC
s==1 p==2
s==2 p==1
-------

...must be considered pure luck or expected result? I mean it in
particular, related to the use of the * and / operators and related to
the fact that the variables get actually swapped.

Francesco

== 5 of 6 ==
Date: Fri, Aug 28 2009 6:19 am
From: Francesco

On 28 Ago, 15:12, Francesco <entul...@gmail.com> wrote:

> [ associativity details snipped, thanks for the clarification ]

> Given the standard specifications, the fact that the following code...
>
> -------
> #include <iostream>
> using namespace std;
> int A(){cout << "A"; return 1;}
> int B(){cout << "B"; return 1;}
> int C(){cout << "C"; return 1;}
> int main()
> {
> A() * B() / C();
> cout << endl;
> A() * (B()) / ((C()));
> cout << endl;
> int s = 1;
> int p = 2;
> cout << "s==" << s << " p==" << p << endl;
> s=s*p/(p=s);
> cout << "s==" << s << " p==" << p << endl;
> return 0;}
>
> -------
>
> ...produces this output...
>
> -------
> ABC
> ABC
> s==1 p==2
> s==2 p==1
> -------
>
> ...must be considered pure luck or expected result? I mean it in
> particular, related to the use of the * and / operators and related to
> the fact that the variables get actually swapped.

Heck, the response was in the part I snipped. Sorry, James.
Hence the above result must be pure luck, swapping included.

Francesco

== 6 of 6 ==
Date: Fri, Aug 28 2009 8:24 am
From: Francesco

On 27 Ago, 12:07, manohar <harirammano...@gmail.com> wrote:
> This is the program of the selection sort, you can see it here
>
> http://pastebin.com/m4338968e
>
> what are the problems i am facing?
>
> 1. Unable to get exact output due to facing problem in recursion that
> too especially when returning the class........
>
> 2.I have checked above program for different values before returning,
> it is good they are in sorted order, but due to recursion property the
> output is mislead.....

Hi manohar, since some problems in your code haven't been pointed out,
I'm taking the occasion to point them out and to resume the thread for
your convenience.

First, this is your code as taken from the link you mentioned:
-------
//selection sort using recursive using class

#include<iostream.h>
#include<conio.h>

class selrec
{
int small,i,n,a[10];

public:
void accept();
selrec selsort(selrec,int );
void display();
}a1;

void selrec::accept()
{
cout<<"\nEnter how many elements: ";
cin>>n;
cout<<"\nEnter the elements: ";
for(i=0;i<n;i++)
cin>>a[i];
}

selrec selrec::selsort(selrec x,int j)
{
int *ptr=&x.a[j];
small=ptr[0];
for(i=1;i<x.n;i++)
if(small>ptr[i])
small=small*ptr[i]/(ptr[i]=small);

ptr[0]=small;
//cout<<" "<<ptr[0];
if(x.n!=1)
{
--x.n;
++j;
selsort(x,j);
}
return x;
}

void selrec::display()
{
cout<<"\nsorted elements are: ";
for(i=0;i<n;i++)
cout<<" "<<a[i];
}

void main()
{
clrscr();
a1.accept();
a1=a1.selsort(a1,0);
a1.display();
}
-------

- Vladimir Jovic suggested you to change the "selsort()" function
because it's needlessly declared as a member function. Either change
the internal implementation or implement it as a separate function.
Vladimir also pointed out a link with which you can get pointers about
the algorithm you want to implement. Read some of the resources
pointed out and rewrite your algo.

- You're swapping the variables with an expression that counts on luck
- that statement is undefined behavior as pointed out by Juha Nieminen
and James Kanze. Add a temporary and do the swap in the right way.

- You've declared a class member "i" that you're using as a local
counter for the "for" loops. Don't do that. Instead, declare it each
time in each "for" loop. The same stands for the other variables.
Declare them in the smallest possible scope and initialize them as
soon as possible - in your case, that would be immediately, in the
declaration.

- You've declared a class member "a[10]" and your "accept()" function
can write outside of its boundaries. Don't use arrays, use
std::vector.

- Read the clc++ FAQ http://www.parashift.com/c++-faq-lite and fix the
other weird things of your code (not-conforming main, superfluous non-
STD headers and function call, standard header included in the ".h"
form and namespace qualifications missing).

Once you have fixed all the above, shall you need again, post the code
and we'll have a look to see what is going wrong - you current
implementation simply doesn't work, I've just compiled it and fed it
"5" as number of elements, "5 4 3 2 1" as elements and it returned "1
5 4 3").

The next time, post the code directly in your message and not
elsewhere on the net, where your code could disappear at any moment,
also give meaningful names to the important variables, so that people
reading your code can understand what they are meant to represent.

You might want to take my points upside-down and start by reading the
FAQ. Take it from the beginning and dig your way through it as much as
you can. You'll learn a lot of things.

Keep on improving your code,
best regards,
Francesco

==============================================================================
TOPIC: Run-time-checking whether a pure virtual function has been implemented
http://groups.google.com/group/comp.lang.c++/t/71c3c8c3041ae524?hl=en
==============================================================================

== 1 of 2 ==
Date: Fri, Aug 28 2009 5:02 am
From: Victor Bazarov

James Kanze wrote:
> On Aug 27, 8:05 pm, Victor Bazarov <v.Abaza...@comAcast.net> wrote:
>> mzdude wrote:
>>> On Aug 27, 9:04 am, Victor Bazarov <v.Abaza...@comAcast.net> wrote:
>>>> Nordlöw wrote:
>>>>> Is it possible to do runtime-checks whether a virtual
>>>>> base-class pure virtual function has been implemented in
>>>>> an inherited class?
>>>> Run-time? Check? Why? Besides, what would be the point?
>>>> If you have the object, it can't be of the abstract base
>>>> type, can it? An object cannot be instantiated with the
>>>> type that is abstract. If you have the pointer to the base
>>>> class, the virtual function table (the usual way to
>>>> implement the mechanism) contains the pointer to the final
>>>> overrider for that function, not the unimplemented pure
>>>> 0...
>
>>>> What problem are you trying to solve?
>
>>> I take to mean
>
>>> #include <iostream>
>>> using namespace std;
>
>>> class Foo {
>>> public:
>>> virtual void bar() = 0 { cout << "Hello "; }
>>> };
>
>>> class Bar : public Foo
>>> {
>>> public:
>>> virtual void bar() { cout << "World\n"; }
>>> };
>
>>> int main()
>>> {
>>> Bar b;
>
>>> // Some kind of test to see if
>>> // Foo::bar is implemented
>
>> You mean there is a definition of that pure virtual function,
>> yes?
>
> I'm not too sure what he is asking, but I don't think it has
> anything to do with this---he did say "implemented in an
> inherited class". (Of course, taken literally, the answer is
> simple: there's no runtime check, because if it isn't
> implemented in a derived class, you can't instantiate objects of
> that type.)

But you can, as subobjects of the some other derived class.
Theoretically, I still might want to check if the "child" has
implemented some pure virtual function or not, which would mean that the
function is defined in some "[great-]grandchild".

>
>>> b.Foo::bar();
>>> b.bar();
>
>>> return 0;
>>> }
>
>> Code that calls a pure virtual function has undefined
>> behavior.
>
> No. Code that would resolve dynamically to a pure virtual
> function is undefined behavior, but code that doesn't use
> dynamic resolution can use it.
>
>> Is that what you're trying to avoid?
>
>> There is no portable way to determine that a pure virtual
>> function has been implemented, AFAICT.
>
> It has been implemented. Otherwise, you can't instantiate
> objects of the type. If the inheritance hierarchy is more than
> two levels, however, there's no way to know whether it was
> instantiated in the most derived class, or in one of the
> intermediate classes. (Maybe that's what he's asking. Maybe
> not.)

Exactly.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

== 2 of 2 ==
Date: Fri, Aug 28 2009 5:49 am
From: Juha Nieminen

Victor Bazarov wrote:
> Juha Nieminen wrote:
>> Victor Bazarov wrote:
>>> Code that calls a pure virtual function has undefined behavior.
>>
>> What do you mean? Why would you be able to write an implementation for
>> a pure virtual function if calling it was UB?
>
> I mean if it has no implementation. Right after this I asked if that
> was what OP was trying to avoid. If it has implementation, there is no
> problem, if it doesn't, the results are undefined. Do I have to spell
> everything?

You were responding to a post where there was a pure virtual function
with an implementation. Thus it sounded like you were saying that
calling it is UB.

==============================================================================
TOPIC: canceling noncopyable feature
http://groups.google.com/group/comp.lang.c++/t/5add783081f15e0e?hl=en
==============================================================================

== 1 of 2 ==
Date: Fri, Aug 28 2009 5:04 am
From: Victor Bazarov

Fraser Ross wrote:
> Will you be suggesting this change to the Boost administrators?

What change do you mean? Please quote enough of the replied-to article,
otherwise the "train of thought" is off the tracks, so to speak.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

== 2 of 2 ==
Date: Fri, Aug 28 2009 6:00 am
From: "Fraser Ross"

"Victor Bazarov"
> Fraser Ross wrote:
>> Will you be suggesting this change to the Boost administrators?
>
> What change do you mean? Please quote enough of the replied-to
> article, otherwise the "train of thought" is off the tracks, so to
> speak.

class noncopyable {
protected:
explicit noncopyable (int) {}
~noncopyable () {}

private:
noncopyable ();
noncopyable (noncopyable const &);
noncopyable & operator=(noncopyable const &);
};

That is the suggestion. It prevents the OP's code compiling at the
expense of having to specifically initialise the base noncopyable.

Fraser.

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==============================================================================
TOPIC: ios_base
http://groups.google.com/group/comp.lang.c++/t/23e730da7f93863c?hl=en
==============================================================================

== 1 of 1 ==
Date: Fri, Aug 28 2009 9:38 am
From: "io_x"

"aegis" <aegis@mad.scientist.com> ha scritto nel messaggio
news:1656a8d0-8186-4d1b-bb19-ac1c84d02a14@t13g2000yqt.googlegroups.com...
> Where is ios_base located?

in a .dll or better in some .obj code

==============================================================================
TOPIC: Need help with printing Unicode! (C++ on CentOS)
http://groups.google.com/group/comp.lang.c++/t/e7dff48651d7be5d?hl=en
==============================================================================

== 1 of 2 ==
Date: Fri, Aug 28 2009 9:51 am
From: Zerex71

Greetings,

I'm sure this has been addressed before but I've hunted all over the
web and no one seems to provide a comprehensive answer. I just want
to do one thing: Under CentOS, in a simple C++ program, I'd like to be
able to print Unicode characters to a console output. For example,
I'd like to print the musical flat, natural, and sharp signs.

Here's what I've done so far:
1. Using Eclipse, created a small C++ console project.
2. Declare three chars, each of type wchar_t, and assigned them their
Unicode values (0x266d, 0x266e, 0x266f).
3. Attempted to print them out using wprintf().
4. Set my output console to a font which can represent the characters
(glyphs?) - Lucida Console

A few observations:
1. I can go to a Unicode code page website and copy the characters
displayed and paste them into my source file which is in the same font
(that was my first trick which ultimately blew me out of the water
because Eclipse was bitching about not being to save the files due to
encoding...tried changing it...then it promptly deleted all my lines
and left me with a bunch of NUL).
2. Mixing cout and wprintf results in the wprintf statements being
totally ignored.
3. Using only wprintf results in "Sign: ?" displayed in the console
output, even though it can display the glyphs correctly when I pasted
them (1.)
4. Calling setlocale() as directed by an example has no effect on my
program.
5. Using fwide() to determine if my setup is legit works because I
don't hit the exit condition that I wrote for that test.

So, I don't know what else to try to get this to work. There's a lot
of stuff about Unicode on Windows out there but I'm not doing Windows,
and figured the Linux community might have an answer.

Thanks.

== 2 of 2 ==
Date: Fri, Aug 28 2009 10:08 am
From: Paavo Helde

Zerex71 <mfeher1971@gmail.com> kirjutas:

> Greetings,
>
> I'm sure this has been addressed before but I've hunted all over the
> web and no one seems to provide a comprehensive answer. I just want
> to do one thing: Under CentOS, in a simple C++ program, I'd like to be
> able to print Unicode characters to a console output. For example,
> I'd like to print the musical flat, natural, and sharp signs.
>
> Here's what I've done so far:
> 1. Using Eclipse, created a small C++ console project.
> 2. Declare three chars, each of type wchar_t, and assigned them their
> Unicode values (0x266d, 0x266e, 0x266f).
> 3. Attempted to print them out using wprintf().
> 4. Set my output console to a font which can represent the characters
> (glyphs?) - Lucida Console

I am not not sure about CentOS, but in Linux generally UTF-8 is used. One
should have an UTF8 locale (e.g. LANG=en_US.utf8). If your code
internally uses wchar_t, then it should be converted to UTF-8 before
output. I am not sure if wprintf() or wcout() can do that automatically.
In our software we use UTF-8 and std::string internally, and it is
working perfectly in Linux.

hth
Paavo

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http://groups.google.com/group/comp.lang.c++/t/5482bd8e63af84d3?hl=en
==============================================================================

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