Would u explain this code about template and class

**Alf P. Steinbach** · Sep 11 '07, 09:25 AM

Re: Would u explain this code about template and class

* Lippman.Gan@gma il.com:

Sorry for that I can not clarify the type of issue in topic.
>
Look the code below:
>
template <class T>
class A
{
public:
class a
{
public:
a(int elem):x(elem){c out<<"a::a()\n" ;}
~a(){cout<<"a:: ~a()\n";}
int GetX(){return x;};
private:
int x;
};
A(a _a1):a1(_a1){co ut<<"haha"<<end l;}
private:
a a1;
friend bool operator ==(const A<int>& t1,const A<int>& t2)
{
cout<<"operator =="<<endl;
return true;
}
>
};
int main()
{
A<intb(10); //This would be OK, why?
if(b==10){cout< <"success\n"; } //Why this can not run?
return 0;
}
>
I see in main()
L1, 10 is an integer it implicitly transfer to class a
L2, 10 is also an integer, but it can not implicitly transfer.
why L1 can do this but L2 can not?

Top-level issue seems to be the declaration of operator== as private.

I haven't compiled the code, so may be more issues.

Cheers, & hth.,

- Alf

**Alf P. Steinbach** · Sep 11 '07, 09:25 AM

Re: Would u explain this code about template and class

* Alf P. Steinbach:

* Lippman.Gan@gma il.com:

>Sorry for that I can not clarify the type of issue in topic.
>>
>Look the code below:
>>
>template <class T>
>class A
>{
> public:
> class a
> {
> public:
> a(int elem):x(elem){c out<<"a::a()\n" ;}
> ~a(){cout<<"a:: ~a()\n";}
> int GetX(){return x;};
> private:
> int x;
> };
> A(a _a1):a1(_a1){co ut<<"haha"<<end l;}
> private:
> a a1;
> friend bool operator ==(const A<int>& t1,const A<int>& t2)
> {
> cout<<"operator =="<<endl;
> return true;
> }
>>
>};
>int main()
>{
> A<intb(10); //This would be OK, why?
> if(b==10){cout< <"success\n"; } //Why this can not run?
> return 0;
>}
>>
>I see in main()
>L1, 10 is an integer it implicitly transfer to class a
>L2, 10 is also an integer, but it can not implicitly transfer.
>why L1 can do this but L2 can not?

>
Top-level issue seems to be the declaration of operator== as private.
>
I haven't compiled the code, so may be more issues.

Uh, wait, I didn't see the "friend" there.

This requires more thinking.

Someone else, not me! (Busy!)

Cheers, & sorry this doesn't help you,

- Alf

**Barry** · Sep 11 '07, 10:35 AM

Re: Would u explain this code about template and class

Lippman.Gan@gma il.com wrote:

Sorry for that I can not clarify the type of issue in topic.
>
Look the code below:
>
template <class T>
class A
{
public:
class a
{
public:
a(int elem):x(elem){c out<<"a::a()\n" ;}
~a(){cout<<"a:: ~a()\n";}
int GetX(){return x;};
private:
int x;
};
A(a _a1):a1(_a1){co ut<<"haha"<<end l;}
private:
a a1;
friend bool operator ==(const A<int>& t1,const A<int>& t2)
{
cout<<"operator =="<<endl;
return true;
}
>
};
int main()
{
A<intb(10); //This would be OK, why?
if(b==10){cout< <"success\n"; } //Why this can not run?
return 0;
}
>
I see in main()
L1, 10 is an integer it implicitly transfer to class a
L2, 10 is also an integer, but it can not implicitly transfer.
why L1 can do this but L2 can not?
>

I guess "no implicit type conversion" in a friend function.

--
Thanks
Barry

**Michal Nazarewicz** · Sep 11 '07, 10:55 AM

Re: Would u explain this code about template and class

(I have modified Lippman's code a bit so that it is a full program which
compiles fine.)

Lippman.Gan@gma il.com writes:

#include <iostream>
>
template <class Tclass A {
public:
class a {
int x;
public:
a(int elem) : x(elem) { std::cout << "a::a()\n"; }
~a() { std::cout << "a::~a()\n" ; }
int GetX() const { return x; }
};
A(a _a1) : a1(_a1) { std::cout << "haha\n"; }
int GetX() const { return a1.GetX(); }
private:
a a1;
};
>
bool operator==(cons t A<int&t1, const A<int&t2) {
std::cout << "operator== \n";
return t1.GetX() == t2.GetX();
}
>
int main() {
A<intb(10); //L1: This would be OK, why?
if (b==10) std::cout << "success\n" ; //L2: Why this can not run?
return 0;
}
>
I see in main()
L1, 10 is an integer it implicitly transfer to class a
L2, 10 is also an integer, but it can not implicitly transfer.
why L1 can do this but L2 can not?

I would say that in L1 there is a direct conversion from int to
A<int>::a (ie. A<int>::a::a(in t) constructor), whereas in L2 to convert
int into A<intyou need two conversions: first convert int into
A<int>::a and then convert it into A<int>. This may be the reason

AFAIK if you want L1 not to compile you can use "explicit" keyword
before a(int) declaration.

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl >---<jid:mina86*chr ome.pl>--ooO--(_)--Ooo--

**Barry** · Sep 11 '07, 11:45 AM

Re: Would u explain this code about template and class

Barry wrote:

Lippman.Gan@gma il.com wrote:
>
I guess "no implicit type conversion" in a friend function.
>
>

My guess is wrong,
Sorry

--
Thanks
Barry

**D. Susman** · Sep 11 '07, 03:15 PM

Re: Would u explain this code about template and class

I guess what is wrong with this code is, in the "==" function you're
trying to implicitly cast an int to type class A. In L1, it is valid
because class a is constructed with an int. Since class A is
constructed with a "class a" variable, implicit conversion request
from int to class A is not sound.
Maybe you anticipated for something like "cascaded implicit type
conversion"? Even if it was implemented as a language feature, this
would lead to being typeless in conversions as hierarchies go deep
which would break down a well typed language.

**James Kanze** · Sep 12 '07, 07:35 AM

Re: Would u explain this code about template and class

On Sep 11, 12:54 pm, Michal Nazarewicz <min...@tlen.pl wrote:

(I have modified Lippman's code a bit so that it is a full program which
compiles fine.)

Lippman....@gma il.com writes:

#include <iostream>

template <class Tclass A {
public:
class a {
int x;
public:
a(int elem) : x(elem) { std::cout << "a::a()\n"; }
~a() { std::cout << "a::~a()\n" ; }
int GetX() const { return x; }
};
A(a _a1) : a1(_a1) { std::cout << "haha\n"; }
int GetX() const { return a1.GetX(); }
private:
a a1;
};

bool operator==(cons t A<int&t1, const A<int&t2) {
std::cout << "operator== \n";
return t1.GetX() == t2.GetX();
}

int main() {
A<intb(10); //L1: This would be OK, why?
if (b==10) std::cout << "success\n" ; //L2: Why this can not run?
return 0;
}

I see in main()
L1, 10 is an integer it implicitly transfer to class a
L2, 10 is also an integer, but it can not implicitly transfer.
why L1 can do this but L2 can not?

I would say that in L1 there is a direct conversion from int to
A<int>::a (ie. A<int>::a::a(in t) constructor), whereas in L2 to convert
int into A<intyou need two conversions: first convert int into
A<int>::a and then convert it into A<int>. This may be the reason

Exactly. In the case of L1, there is an implicit conversion int
to A<int>::a, and the A<int>::a is used to initialize the
variable. One user defined conversion, so OK. In the case of
L2, we need an A<int>, and it would require two user defined
conversions (int to A<int>::a, then A<int>::a to A<intto get
there), so it is illegal. Note that:

A<intb = 10 ;

would be illegal too, since copy initialization requires
converting the initialization expression to the target type
(here, A<int>).

AFAIK if you want L1 not to compile you can use "explicit" keyword
before a(int) declaration.

And if he wants L2 to compile, he should add a constructor to A
which takes a T.

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

Would u explain this code about template and class

Would u explain this code about template and class

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