🚀 go-pugleaf

>>>>> "RL" == Roman Lechtchinsky <wolfro@cs.tu-berlin.de> writes:
[...]
>> If a constructor's parameters all have default values, it can be a
>> converting constructor, and declaring it "explicit" prevents it from
>> being invoked implicitly.

RL> Yes, I've had this in mind, too. In the following case:

RL> class A
RL> {
RL>  explicit A( int x=0 );
RL> };

RL> how do you call the default constructor? Can it still be invoked
RL> implicitly and what is the "explicit" syntax if not? IMO, something like

RL> A a;

RL> is not explicit ( maybe I'm wrong here and this does explicitly call the 
RL> default constructor explicitly - but where does the DWP define it? ). An 
RL> explicit call would rather be

Yes: the above is quite ``explicit''. Implicit would be:

	a = 23; // Implicitly equivalent to `a = A(23)' if no `explicit'
		// specification (and no preferred int-assignment
		// alternative).

RL> A a();

This is not a constructor invacation anyway (it's the declaration of
a function `a' returning an object of type `A').

	Daveed


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