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ferdinand.stefanus@gmail.
Guest
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 Posted: Tue Dec 21, 2004 2:42 am Post subject: Templated class constructor question |
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Hi, I have some questions regarding templated class constructor:
#include <iostream>
using namespace std;
template<typename T>
class Foo
{
public:
explicit Foo(const T& bar): _bar(bar)
{ cout << "In constructor, _bar = " << _bar << 'n'; }
template
Foo(const Foo<U>& rhs): _bar(rhs.GetBar())
{ cout << "In converting constructor, _bar = " << _bar << 'n'; }
Foo(const Foo
{ cout << "In copy constructor, _bar = " << _bar << 'n'; }
T GetBar() const { return _bar; }
private:
T _bar;
};
int main()
{
Foo
Foo<int> intFoo3(intFoo1);
Foo<double> dblFoo1(intFoo2);
}
A quick test with VC++6 yields the following result:
In constructor, _bar = 1
In constructor, _bar = 2
In copy constructor, _bar = 1
In converting constructor, _bar = 2
Is this behaviour conforming with the standard? It seems that even if I
remove the copy constructor declaration/definition, the converting
constructor will not be used (the compiler-generated copy constructor
will be used instead). If that's the case, can I safely assume that the
converting constructor will never be called for templated class of the
same type?
Thanks!
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Victor Bazarov
Guest
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| Posted: Tue Dec 21, 2004 4:04 am Post subject: Re: Templated class constructor question |
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<ferdinand.stefanus (AT) gmail (DOT) com> wrote...
| Quote: | Hi, I have some questions regarding templated class constructor:
#include <iostream
using namespace std;
template
class Foo
{
public:
explicit Foo(const T& bar): _bar(bar)
{ cout << "In constructor, _bar = " << _bar << 'n'; }
template
Foo(const Foo
{ cout << "In converting constructor, _bar = " << _bar << 'n'; }
Foo(const Foo
{ cout << "In copy constructor, _bar = " << _bar << 'n'; }
T GetBar() const { return _bar; }
private:
T _bar;
};
int main()
{
Foo
Foo<int> intFoo3(intFoo1);
Foo<double> dblFoo1(intFoo2);
}
A quick test with VC++6 yields the following result:
In constructor, _bar = 1
In constructor, _bar = 2
In copy constructor, _bar = 1
In converting constructor, _bar = 2
Is this behaviour conforming with the standard?
|
Yes, AFAICS.
| Quote: | It seems that even if I
remove the copy constructor declaration/definition, the converting
constructor will not be used (the compiler-generated copy constructor
will be used instead).
|
Yes, that's correct.
| Quote: | If that's the case, can I safely assume that the
converting constructor will never be called for templated class of the
same type?
|
Yes, that's the Standard requirement, IIRC. I couldn't find the passage
from the Standard within 5 minutes, though...
V
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Jonathan Mcdougall
Guest
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| Posted: Tue Dec 21, 2004 7:40 am Post subject: Re: Templated class constructor question |
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[email]ferdinand.stefanus (AT) gmail (DOT) com[/email] wrote:
| Quote: | Hi, I have some questions regarding templated class constructor:
#include <iostream
using namespace std;
template
class Foo
{
public:
explicit Foo(const T& bar): _bar(bar)
{ cout << "In constructor, _bar = " << _bar << 'n'; }
template
Foo(const Foo
{ cout << "In converting constructor, _bar = " << _bar << 'n'; }
Foo(const Foo
{ cout << "In copy constructor, _bar = " << _bar << 'n'; }
T GetBar() const { return _bar; }
private:
T _bar;
};
int main()
{
Foo
Foo<int> intFoo3(intFoo1);
Foo<double> dblFoo1(intFoo2);
}
A quick test with VC++6 yields the following result:
In constructor, _bar = 1
In constructor, _bar = 2
In copy constructor, _bar = 1
In converting constructor, _bar = 2
Is this behaviour conforming with the standard?
|
Yes.
| Quote: | It seems that even if I
remove the copy constructor declaration/definition, the converting
constructor will not be used (the compiler-generated copy constructor
will be used instead).
|
Templated copy-ctor, ctor and assignment operator never replace
non-templated ones, so even if you declare them, the compiler will still
generate default ones.
| Quote: | If that's the case, can I safely assume that the
converting constructor will never be called for templated class of the
same type?
|
Yes.
Jonathan
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