Return pointer-to-member in const method

jsulm

@JonB said in Return pointer-to-member in const method:

int *pointerToMember() const { return &member; }

const int *pointerToMember() const { return &member; }

Else you would be able to change the state of the object using the pointer even though the method is marked as const.

JonB

@jsulm
But I know this, which is why I wrote

so I don't want to return const int *, but this method itself does not alter member

I was asking (nicely) if you have a reference to this aspect of const method in C++ I could read up on? I do understand a const method cannot alter a member, but in itself this method does not do so.

Let me try this: sometimes I use this return result just for read, sometimes I want to write back into the pointed-to. Obviously if I just remove the const off the method, I cannot call it in the read-only case from some other class const method.

How to achieve? Do I have to provide two methods:

int *pointerToMember() { return &member; }
const int *pointerToMember() const { return &member; }

I have a feeling you may answer "yes" to that?

Supposing you do: assume the body of the method is several lines long. [For the record, it's actually a look-up in a member array of structs, and I want to return a pointer to the found element/nullptr, not an index.] So I do not want to have to re-write the code in each of these two methods. How can I write it only once? I'm going to have trouble if my non-const overload tries to call the const overload, or if the const overload tries to call the non-const one, either way, aren't I?

jsulm

@JonB said in Return pointer-to-member in const method:

How can I write it only once?

You can call the const version in non-const version and const_cast the return value from the const version. const_cast should of course be avoided, but in your scenario I think it is OK.

const int *pointerToMemberConst() const { return &member; }
int *pointerToMember()  { return const_cast<int*>(pointerToMemberConst()); }

I don't have a good source for this topic at hand.

JonB

@jsulm
Yes, and thank you, that was all I could think of doing myself, that way round. It's just that borrowing from @Chris-Kawa over at https://forum.qt.io/topic/120328/std-vector-qvector-and-fields/11, his

and a cute fluffy kitten will gruesomely die somewhere every time you do that :(

applies to (de-)const-casting ;-)

jsulm

@JonB Add a "goto" to make him even more "happy" :-)

JonB

@jsulm
OK, we're nearly there, but we have a wrinkle.

I want these two methods to have the same name. So I don't want the Const suffix on the const one.

I am assuming we can do this in C++, because Qt has methods like QVector:

T &	operator[](int i)
const T &	operator[](int i) const

(I know that's an operator, but I think if I looked around there are other methods which have both const & non-const variants?)

So I try:

const int *pointerToMember() const { return &member; }
int *pointerToMember()  { return const_cast<int*>(pointerToMember()); }

But the second overload gives Creator yellow-triangle clang warning

warning: all paths through this function will call itself

gcc compiles without warning. But when I test at run-time, sure enough, I get infinite recursion: the second overload calls itself, not the first one, and I am foo-barred :(

So.... Qt seems to manage with same-named const & non-const variants. How can I adapt your suggestion, so they use common code yet work as required, please? E.g. is there anything I can write here to make the second overload's pointerToMember() call the first one, not itself, some "overloadOf" or something?

EDIT
I guess I could do:

private:
    const int *_pointerToMember() const { return &member; }

public:
    const int *pointerToMember() const { return _pointerToMember(); }
    int *pointerToMember()  { return const_cast<int*>(_pointerToMember()); }

Can I do it without that extra dummy method? Neater, and guaranteed [don't start on me with inline ;-) ] not to have an inefficient extra function call :)

jsulm

@JonB It's getting ugly :-)

const int *pointerToMember() const { return &member; }
int *pointerToMember()  { const MyClass *_this = this;  return const_cast<int*>(_this->pointerToMember()); } // Now compiler knows that you want to call const pointerToMember

JonB

@jsulm
Hmm, good one, thanks! Now I have an extra assignment to slow my code down :( ;-)

This is indeed fine, and I get it. Nonetheless, final question: is there something in C++ to do with like "overload of" which can pick between the const & non-const variants here? Just because I am interested. You know that when we use Qt connect() we can use QOverload<...>::of to pick desired function overloads, I have a feeling that resolves to some C++ "overload of" statement? But here we are trying to distinguish between signatures which do not differ in parameters, I'm trying to understand whether that can be done?

fcarney

const auto value1 = obj->pointerToMember(); // should be const version
auto value2 = obj->pointerToMember(); // should be non-const version

JonB

@fcarney

    const int *pointerToMember() const { ... }
    int *pointerToMember()  { ... }

const auto value1 = obj->pointerToMember(); // should be const version

Nope (but thanks for your interest!), that const one steps straight into the second, non-const variant! (As does the other call too.)

I would not expect const variable = method() to call the const variant of the method, particularly. So far as I know, that const on the variable perfectly allows it to pick any non-const method, it only chooses to treat the return result as const but no other effect. Which is what it does it here.

fcarney

Interesting, maybe don't overload then. I tried as reference to see if it makes a difference. No change.

JonB

@jsulm said in Return pointer-to-member in const method:

int *pointerToMember()  { const MyClass *_this = this;  return const_cast<int*>(_this->pointerToMember()); } // Now compiler knows that you want to call const pointerToMember

I wrote

Now I have an extra assignment to slow my code down :( ;-)

So I adapted, to remove _this:

int *pointerToMember()  { return const_cast<int*>( const_cast<const MyClass *>(this)->pointerToMember() ); }

I prefer yours for readability :)

Chris Kawa

I left you for one minute and there's const_cast and goto :P

Getting back to original problem. At first glance you could do something like this:

class Foo
{
private:
    std::array<Stuff, 42> data;
    int indexOf(Key key) const { return /* some lengthy way to determine the index */; };
 
public:
    Stuff* get(Key key) { int index = indexOf(key); return (index < 0) ? nullptr : &data[index]; }
    const Stuff* get(Key key) const { int index = indexOf(key); return (index < 0) ? nullptr : &data[index]; }
};

no casts but I would argue that this is a patch work. The solution becomes a lot nicer if you dig into the problem, which is you're trying to pack two things into one getter - a logic to determine if given element is present and retrieving it. Those are two things and they also incur a performance cost (branching) so I'd say design your interface so that the decision about taking the cost or not is left to the user of your class:

class Bar
{
private:
    std::array<Stuff, 42> data;
    
public:
    int indexOf(Key key) const { return /* some lengthy way to determine the index */; };
    Stuff& at(int index) { return data[index]; }
    const Stuff& at(int index) const { return data[index]; }
};

Shorter, easier, faster and class doesn't absorb responsibilities it doesn't need to. Also those at methods will most definitely get inlined and disappear (inlining is real and super important, don't dismiss it :) )

As for the other thing:

const auto value1 = obj->pointerToMember(); // should be const version
auto value2 = obj->pointerToMember(); // should be non-const version

const or non-const is determined by the constness of the object it is called on, not unrelated variable the result is assigned to, so:

auto value1 = const_obj->pointerToMember(); // const version, auto resolves to const something
auto value2 = obj->pointerToMember(); // non-const version, auto resolves to non-const something

Btw. this is a source of a common performance trap with Qt and C++11:

QVector<Stuff> stuff;

for (Stuff& foo : stuff) {}  //no! calls non-const begin/end and can cause expensive detach()

so people think "oh, I should just add const and it's fine":

for (const Stuff& foo : stuff) {}  //no! still calls non-const begin/end

The proper way to do it is:

for (const Stuff& foo : qAsConst(stuff)) {}  //ok, calls const begin/end

JonB

@Chris-Kawa said in Return pointer-to-member in const method:

I left you for one minute and there's const_cast and goto :P

That's what happens to fluffy kittens if you turn your back....

Let's pick my one of what (I understand) you suggest:

    int indexOf(Key key) const { return /* some lengthy way to determine the index */; };
    Stuff& at(int index) { return data[index]; }
    const Stuff& at(int index) const { return data[index]; }

So, you avoid the problem by making the lookup function return an int index. Then you return const/non-const data[index].

This breaks my (unspoken) "efficiency" criterion! My lookup marches through the array without an int counter, only with a pointer (it can return nullptr on not-found, so no references here), and returns that. You will make me turn that into an index, and then you will turn it back by adding it onto data.

I am shocked. I was speaking to someone recently in another thread here, and they berated me for the overhead of indexing into arrays, when I said it was "milliseconds" they said "every microsecond counts, in game development, this is not to be ignored". Can you think who that was? :D

Chris Kawa

@JonB said:

Can you think who that was? :D

Yup, and I stand by what I said, which was not indexing into an array but indexing into a vector. Huge huge difference.

The int index is just an example. Every case is different and it doesn't have to be an int. It could be an iterator, some hash thingie or whatever is most efficient in your case. Pointer has that nasty proprty of being both index and data at the same time, which causes your problems - you want a const pointy thing and a non-const data thing. One variable can't be both at the same time so that's why I'm suggesting to split them up.

JonB

@Chris-Kawa

int indexOf(Key key) const { return /* some lengthy way to determine the index */; };

The problem here is, that function returning int does not exhibit the problem! You don't have to worry about returning int versus const int. It's when function returns a something *. int method() const is never a problem, but int *method() const where it returns a member variable is a problem. So I see I then need const int *method() const as well as int *method(). Once your indexOf() returns a pointer into a member variable thingie you end up needing a const something *indexOf() const and a something *indexOf(Key key), for a method which does the same thing.

Anyways. My head is hurting on this. We may all be saying the same thing in different ways.

Chris Kawa

@JonB said in Return pointer-to-member in const method:

The problem here is, that function returning int does not exhibit the problem!

Exactly, it's intentional on my part and that's the point. Don't try to solve an ugly problem. Redesign and untangle dependencies so there is no problem in the first place ;) int* is an "index" and int* points to data. Untangle those roles.

Anyways. My head is hurting on this.

I fear I might have accidentally terrorized you into being paranoid about something that will bring you marginal gains and make your code a lot worse to read/maintain. I'd say, just for test, do the duplicated const and non-const methods, measure how much gain are you actually getting, decide if it's worth it and only then proceed or revert. While I care deeply about performance there is a line below which it's just not worth it, as in how much optimizations can you achieve in a finite amount of time and how it reflects on readability and ease of maintenance. I'd just like that line to be lower than it usually is, but it's up to you really. Don't let me pressure you too much. I've been told I can be bossy ;)

JonB

@Chris-Kawa said in Return pointer-to-member in const method:

measure how much gain are you actually getting, decide if it's worth it and only then proceed or revert

Sounds like what I would say :)

While I care deeply about performance

I do, but kinda more just in an algorithmic sense than whether it makes any visible difference to what I'm doing.

Don't let me pressure you too much. I've been told I can be bossy ;)

Not at all! I read your posts with interest, high quality.

This has all revealed something to me which I had not appreciated. I thought Class::method() const only guaranteed that it did not alter *this. I did not expect that, for safety, it also does not allow Class::Member *Class::method() const. That function does not itself alter *this, but I do see that it returns a write-pointer into const this * which could later be used to do so. Hence you have to make that return a const * if you want method() const I just was not aware of this.

I'm sure there are just pages of C++ specs I could/ought to read up on const.... [Actually I think I did so a while ago, I recall it being longggggg.]

P.S.
When I started C it didn't have const yet. Lambs gambolled carefree in the fields, life was easy then...

Christian Ehrlicher

If you want some further discussion points, take a look at how the Qt api returns pointers:

QLayoutItem *QGridLayout::itemAtPosition(int row, int column) const
QUndoStack *QUndoGroup::activeStack() const
QObject *QDropEvent::source() const

want more? :)

Chris Kawa

const is kinda like the GPL license - infectious and intentionally so :)
I won't bring the actual standardese page number, but a close-enough rule for this is on cppreference:

If the operand is an lvalue expression of some object or function type T, operator& creates and returns a prvalue of type T*, with the same cv qualification, that is pointing to the object or function designated by the operand.

What this piece of the typical standardese mumbo jumbo translates to is that when you're doing &member inside a const method it's really &(this->member) and cv-qualifiers (const and volatile) for the resulting pointer are taken from the object this points to. Since you're inside a const method this points to a const object in that scope and so & returns a pointer to const member.

Btw. I've seen an interesting debate somewhere (can't find it now, it was a while ago) about if this should be a const pointer to const object or just a pointer to const object i.e. T const * vs T const * const. The argument for non-const this pointer was some wizardry with modifying this inside a member to avoid vtables. It landed on this being a non-const prvalue and thus non-assignable, but those are some deep trenches :)

@Christian-Ehrlicher said in Return pointer-to-member in const method:

If you want some further discussion points, take a look at how the Qt api returns pointers:

I believe the last two are just straight pointer retrievals so not a big deal. The first one needs that branching logic I mentioned so it's basically against all I've said, but that's a design choice Qt takes. It is well known to take small performance hits here and there for the sake of ease of use and I think it's a fair compromise for all that it offers in return - consistency being a big one. Not a design I would make but hey, can't have it all the way I like :)