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Constructor with default argument
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Daniel
Oct 10, 2018, 6:01:07 PM10/10/18
to
Consider the following:
class A
{
int n;
public:
A() = delete;
A(int n)
: n(n)
{
}
};
template <class T>
class B
{
T t;
public:
B(T t = T())
: t(t)
{
}
};
(1) B<A>(); // fails, as expected
(2) B<A>(A(1)); // ?
Is it for certain that the expression T() won't get evaluated in (2)?
Thanks,
Daniel
class A
{
int n;
public:
A() = delete;
A(int n)
: n(n)
{
}
};
template <class T>
class B
{
T t;
public:
B(T t = T())
: t(t)
{
}
};
(1) B<A>(); // fails, as expected
(2) B<A>(A(1)); // ?
Is it for certain that the expression T() won't get evaluated in (2)?
Thanks,
Daniel
Öö Tiib
Oct 11, 2018, 1:51:52 AM10/11/18
to
On Thursday, 11 October 2018 01:01:07 UTC+3, Daniel wrote:
> Consider the following:
>
> class A
> {
> int n;
> public:
> A() = delete;
> A(int n)
> : n(n)
> {
> }
> };
>
> template <class T>
> class B
> {
> T t;
> public:
> B(T t = T())
> : t(t)
> {
> }
> };
>
> (1) B<A>(); // fails, as expected
You mean by being most vexing parse?
> Consider the following:
>
> class A
> {
> int n;
> public:
> A() = delete;
> A(int n)
> : n(n)
> {
> }
> };
>
> template <class T>
> class B
> {
> T t;
> public:
> B(T t = T())
> : t(t)
> {
> }
> };
>
> (1) B<A>(); // fails, as expected
> (2) B<A>(A(1)); // ?
>
> Is it for certain that the expression T() won't get evaluated in (2)?
Öö Tiib
Oct 11, 2018, 1:56:15 AM10/11/18
to
On Thursday, 11 October 2018 08:51:52 UTC+3, Öö Tiib wrote:
> > (1) B<A>(); // fails, as expected
>
> You mean by being most vexing parse?
Nah ... mea culpa ... it isn't most vexing parse.
> > (1) B<A>(); // fails, as expected
>
> You mean by being most vexing parse?
Fraser Ross
Oct 11, 2018, 4:40:08 AM10/11/18
to
On 10/10/2018 23:00, Daniel wrote:
> Consider the following:
>
> class A
> {
> int n;
> public:
> A() = delete;
> A(int n)
> : n(n)
> {
> }
> };
The class has a constructor so the default constructor is not implicitly
> Consider the following:
>
> class A
> {
> int n;
> public:
> A() = delete;
> A(int n)
> : n(n)
> {
> }
> };
declared defaulted and hence you did not need to declare it deleted.
Its an old rule but the terminology has changed.
Fraser.
Daniel
Oct 11, 2018, 9:58:36 AM10/11/18
to
On Thursday, October 11, 2018 at 1:51:52 AM UTC-4, Öö Tiib wrote:
> On Thursday, 11 October 2018 01:01:07 UTC+3, Daniel wrote:
> >
> > class A
> > {
> > int n;
> > public:
> > A() = delete;
> > A(int n)
> > : n(n)
> > {
> > }
> > };
> >
> > template <class T>
> > class B
> > {
> > T t;
> > public:
> > B(T t = T())
> > : t(t)
> > {
> > }
> > };
> >
> > (1) B<A>(); // fails, as expected
>
> You mean by being most vexing parse?
>
No, this fails because the template argument supplied doesn't have a default
> On Thursday, 11 October 2018 01:01:07 UTC+3, Daniel wrote:
> >
> > class A
> > {
> > int n;
> > public:
> > A() = delete;
> > A(int n)
> > : n(n)
> > {
> > }
> > };
> >
> > template <class T>
> > class B
> > {
> > T t;
> > public:
> > B(T t = T())
> > : t(t)
> > {
> > }
> > };
> >
> > (1) B<A>(); // fails, as expected
>
> You mean by being most vexing parse?
>
constructor, so T() isn't defined.
> > (2) B<A>(A(1)); // ?
> >
> > Is it for certain that the expression T() won't get evaluated in (2)?
>
> Yes.
the default parameter (and hence fail) if an actual parameter was passed.
Daniel
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