Thread: C++ question

Are you trying to say that a destructor isn't a function because it doesn't return anything?

I don't have a book with me, but I'm pretty sure that Stroustrup refers to constructors and destructors as functions despite the fact that they don't have return values.

EDIT: Yup, they're still called member functions -- from Stroustrup

Note update on previous post

Originally posted by elad
>>Because pure virtual destructors must have a definition/implementation, so "C" isn't really correct.

The above is incorrect. Below is an example of a pure virtual function. Answer C is the correct answer to the question.

A virtual function is made pure by initializing it with zero, as in

virtual void Draw() = 0;

Then you've never used a pure virtual destructor

A pure virtual destructor uses the = 0 notation and makes it so you can't create objects from the class (it makes the class abstract) however you still must provide a definition for the function.

Look it up.

because it doesn't say there whether or not a destructor is considered a function, however, previously your claim was that it wasn't a function because it had no return type yet Stroustrup specifically calls constructors functions. I posted the response prior to you bringing up the fact of them not being able to be overloaded, which is a good point, though I don't think it makes them not functions. They're just functions which can't be overloaded. If anyone has a book by Stroustrup handy then we'd be able to get a better answer.

They can be explicitly called, they can be virtual, they can be defined by the user.

Originally posted by Eibro
Exactly, they're destructors are referred to as class members.

That doesn't mean it's not a member function. He just doesn't go on to say anything else.

Originally posted by Eibro
Read into it all you want, I simply asked what data type a destructor returned. I never said a destructor wasn't a function, based souly on the fact that it doesn't return a value.

Well then what was your point in the statement. That's the only point I can think of which would have any relevance to the prior post.

Originally posted by Eibro
Lets look at the definition of a function:

A destructor fits that definition.

C++ Standard 12.4.7:

A destructor can be declared virtual (10.3) or pure virtual (10.4); if any objects of that class or any derived class are created in the program, the destructor shall be defined. If a class has a base class with a virtual destructor, its destructor (whether user or implicitly declared) is virtual.

Find the error(s) in each of the following and explain how to correct it (them).
a) Assume the following prototype is declared in class Time:

Code:

void ~Time( int )

b) The following is a partial definition of class Time.

Code:

class Time {
public:
   // function prototypes
private:
   int hour = 0;
   int minute = 0;
   int second = 0;
};

c) Assume the following prototype is declared in class Employee.

Code:

int Employee( const char *, const char * );

In what context might the name operator/ be used in C++?


How does the precedence of an overloaded operator in C++ compare with the precedence of the original operator?


Name the 5 C++ operators that cannot be overloaded.

Originally posted by Eibro
And because you saw the word "function" in the definition of a constructor doesn't mean a destructor is a function.

I posted that in the response to you apparently claiming that it wasn't a function because it had no return value. A constructor has no return value but it's a function so I posted that as proof that not all functions have a return type. No, it doesn't specifically say there that a destructor is a function but it contradicted your prior post which implied that functions needed a return type (though now you are apparently saying that that is not what you meant).

Originally posted by Eibro
I asked you a question, I didn't make a statement. You wanted to get picky about things, not me.

So you're saying that you made a post which had absolutely no relevance what-so-ever to the topic? I see

Originally posted by Eibro
No, it doesn't.

Yes it does -- it fits the definition completely.

Nowhere in the C++ Standard does it say that destructors are member functions, but:

"The invocation of a destructor is subject to the usual rules for member functions"

IMO, they are functions, a special kind of functions.