Quick Q: May a destructor be final?

Quick A: Yes, as any other virtual member function.

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May a destructor be final?

May a C++ destructor be declared as final?


And if so, does that prevent declaration of a derived class:

Yes, because the derived class would have to declare a destructor (either explicitly by you or implicitly by the compiler), and that destructor would be overriding a function declared final, which is ill-formed.

The rule is [class.virtual]/4:

If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from B a function D​::​f overrides B​::​f, the program is ill-formed.

It's the derivation itself that is ill-formed, it doesn't have to be used.

Is declaring a destructor to be final a workable idiom for indicating that a class is not intended to be used as a base class?

Effectively, but you should just mark the class final. It's quite a bit more explicit.

How to Use the STL With Legacy Output Collections—Jonathan Boccara

And how back_inserter works.

How to Use the STL With Legacy Output Collections

by Jonathan Boccara

From the article:

When you start using the STL and its algorithms in your code, it’s a bit of a change of habits. And then after a while you get used to it. Then it becomes a second nature. And then even your dreams become organized into beautifully structured ranges that fly in and out of well-oiled algorithms.

And when you reach that point, there is no coming back.

Until the day you come upon an old legacy structure that won’t let itself approached by the elegant and expressive way of coding that STL algorithms have. It’s a terrible encounter, where the beast tries to suck you back into the lengthy and dangerous quicksand of the raw for loops that now seemed so far away...

Simplifying Compile-Time Options With if constexpr—Philippe Groarke

New possibilities open in front of us!

Simplifying Compile-Time Options With if constexpr

by Philippe Groarke

From the article:

My latest little experiment relates to compile-time options and eliminating preprocessor checks in user code. I’m not a big fan of MACROs, especially when they are simply used to make compile-time branches. I am also not a fan of other techniques used to minimize this problem. With C++17, we now have a beautiful and simple tool that can help remove all these preprocessor checks, if constexpr...

Quick Q: What is the usefulness of `enable_shared_from_this`?

Quick A: It allows you to get a valid shared pointer from your instance directly.

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What is the usefulness of `enable_shared_from_this`?

It enables you to get a valid shared_ptr instance to this, when all you have is this. Without it, you would have no way of getting a shared_ptr to this, unless you already had one as a member. This example from the boost documentation for enable_shared_from_this:

class Y: public enable_shared_from_this<Y>

    shared_ptr<Y> f()
        return shared_from_this();

int main()
    shared_ptr<Y> p(new Y);
    shared_ptr<Y> q = p->f();
    assert(p == q);
    assert(!(p < q || q < p)); // p and q must share ownership

The method f() returns a valid shared_ptr, even though it had no member instance. Note that you cannot simply do this:

class Y: public enable_shared_from_this<Y>

    shared_ptr<Y> f()
        return shared_ptr<Y>(this);

The shared pointer that this returned will have a different reference count from the "proper" one, and one of them will end up losing and holding a dangling reference when the object is deleted.

enable_shared_from_this is going to be a part of the new C++0x standard as well, so you can also get it from there as well as from boost.

Abstraction design and implementation: `repeat`—Vittorio Romeo

This series of two articles covers the train of thought behind the design and implementation of a simple `repeat` abstraction that, given a number `n` and a function object `f`, invokes `f` `n` times. The abstraction can be used to repeat actions both at run-time and compile-time. The articles also cover importance of propagating `noexcept`-correctness (and the pain caused by it!).

abstraction design and implementation: `repeat`

compile-time `repeat` & `noexcept`-correctness

by Vittorio Romeo

From the articles:

In my previous "passing functions to functions" and "zero-overhead C++17 currying & partial application" articles I've praised C++11 (and newer standards) for allowing us to write "more functional" code. [...] If I want to repeat an action n times, I exactly want to write that in my code:

    repeat(10, []

Cannot get simpler than that - let's implement it!

Quick Q: can i use move only exception throwable objects with vectors?

Quick A: Yes, but with unspecified behavior in case of exception thrown.

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Using an object without copy and without a noexcept move constructor in a vector. What actually breaks and how can I confirm it?

A vector reallocation attempts to offer an exception guarantee, i.e. an attempt to preserve the original state if an exception is thrown during the reallocation operation. There are three scenarios:

  1. The element type is nothrow_move_constructible: Reallocation can move elements which won't cause an exception. This is the efficient case.
  2. The element type is CopyInsertable: if the type fails to be nothrow_move_constructible, this is sufficient to provide the strong guarantee, though copies are made during reallocation. This was the old C++03 default behaviour and is the less efficient fall-back.
  3. The element type is neither CopyInsertable nor nothrow_move_constructible. As long as it is still move-constructible, like in your example, vector reallocation is possible, but does not provide any exception guarantees (e.g. you might lose elements if a move construction throws).

The normative wording that says this is spread out across the various reallocating functions. For example, [vector.modifiers]/push_back says:

If an exception is thrown while inserting a single element at the end and T is CopyInsertable or is_nothrow_move_constructible_v<T> is true, there are no effects. Otherwise, if an exception is thrown by the move constructor of a non-CopyInsertable T, the effects are unspecified.

I don't know what the authors of the posts you cite had in mind, though I can imagine that they are implicitly assuming that you want the strong exception guarantee, and so they'd like to steer you into cases (1) or (2).

Mutable—Arne Mertz

Do you know that keyword?


by Arne Mertz

From the article:

The mutable keyword seems to be one of the less known corners of C++. Yet it can be very useful, or even unavoidable if you want to write const-correct code or lambdas that change their state...

Quick Q: Why would one use nested classes in C++?

Quick A: To hide implementation details

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Why would one use nested classes in C++?

Nested classes are cool for hiding implementation details


class List
        List(): head(NULL), tail(NULL) {}
        class Node
                  int   data;
                  Node* next;
                  Node* prev;
        Node*     head;
        Node*     tail;

Here I don't want to expose Node as other people may decide to use the class and that would hinder me from updating my class as anything exposed is part of the public API and must be maintained forever. By making the class private, I not only hide the implementation I am also saying this is mine and I may change it at any time so you can not use it.

Look at std::list or std::map they all contain hidden classes (or do they?). The point is they may or may not, but because the implementation is private and hidden the builders of the STL were able to update the code without affecting how you used the code, or leaving a lot of old baggage laying around the STL because they need to maintain backwards compatibility with some fool who decided they wanted to use the Node class that was hidden inside <list>.