What is operator T* (where T is a template parameter ) in C++?

Question

class NullClass{
    public:
    template<class T>
        operator T*() const {return 0;}

};

I was reading Effective C++ and I came across this class, I implemented the class and it compiles. I have a few doubts over this:

It doesn't have a return type.
What is this operator.
and what it actually does.

Perhaps a duplicate of something like this: http://stackoverflow.com/questions/1307876/how-do-conversion-operators-work-in-c — Bill Lynch, Jul 31 '14 at 13:44
It's a [used-defined conversion operator](http://en.cppreference.com/w/cpp/language/cast_operator). It allows instances of `NUllClass` to be implicitly converted to a pointer of any type. — Some programmer dude, Jul 31 '14 at 13:44

Quentin · Accepted Answer · 2014-08-01T14:54:55.543

15

That's the type conversion operator. It defines an implicit conversion between an instance of the class and the specified type (here T*). Its implicit return type is of course the same.

Here a NullClass instance, when prompted to convert to any pointer type, will yield the implicit conversion from 0 to said type, i.e. the null pointer for that type.

On a side note, conversion operators can be made explicit :

template<class T>
explicit operator T*() const {return 0;}

This avoid implicit conversions (which can be a subtle source of bugs), but permits the usage of static_cast.

edited Aug 01 '14 at 14:54

answered Jul 31 '14 at 13:45

Quentin

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Not only can, but certainly also should be made explicit. – Deduplicator Jul 31 '14 at 13:52
1

@Deduplicator we're precisely in the case where it should be implicit. You wouldn't want to keep `static_cast`ing `nullptr`, would you ? However I agree that implicit cast operators (and constructors) should be thought about twice. – Quentin Jul 31 '14 at 14:00
Ah, you said "such conversion operators". My comment was more for the general case, though in this specific instance (a deficient re-implementation of `nullptr`), you are sure right that implicit is better. No rules without exceptions (even this rule has a few ;-)). – Deduplicator Jul 31 '14 at 14:03
@Deduplicator I've dropped that "such" because I don't want to define it. Deciding of the explicitness of functions is a whole topic on its own, and outside the scope of the question, though you're right about using implicits sparsely :) – Quentin Jul 31 '14 at 14:08
What is an "implicit cast"? – Lightness Races in Orbit Jul 31 '14 at 19:02
@LightnessRacesinOrbit "Implicit cast" is a sloppy synonym for "coercion". – fredoverflow Jul 31 '14 at 19:03
@FredOverflow: Very sloppy, since "cast" is a synonym for "explicit conversion". – Lightness Races in Orbit Jul 31 '14 at 19:26
1

@LightnessRacesinOrbit Agreed. It's like an *expensive bargain* or a *hornless unicorn* ;) – fredoverflow Aug 01 '14 at 09:35
1

@LightnessRacesinOrbit and what single word would you pedantic geniuses use to name all of converting, reinterpreting, casting and so on ? So these comments can be, like, productive or something. – Quentin Aug 01 '14 at 09:47
@Quentin: Those are _conversions_. – Lightness Races in Orbit Aug 01 '14 at 12:52
@LightnessRacesinOrbit does this apply to reinterpretation (punning) as well ? – Quentin Aug 01 '14 at 12:56
@Quentin: Yes. `[expr.reinterpret.cast]` uses the term "convert" in basically every paragraph. – Lightness Races in Orbit Aug 01 '14 at 14:29
@LightnessRacesinOrbit Thanks for the information. I've edited away that "implicit cast" (which is actually a francicism). – Quentin Aug 01 '14 at 14:57

What is operator T* (where T is a template parameter ) in C++?

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