Using a std::remove_if for vector of non-pointer types

Question

I have a vector of objects :

std::vector<QueuedMessage> _messages;

And trying to remove all objects with erase-remove idiom:

void MessageQueue::removeFromQueue(){
    _messages.erase(std::remove_if(_messages.begin(),_messages.end(),wasMessageSend),_messages.end());
}

My compare function:

bool MessageQueue::wasMessageSend(const QueuedMessage& mess){

            return mess.canSendMessage();
        }

And compare function inside the objects class:

bool QueuedMessage::canSendMessage() const{
    return (SDL_GetTicks() >= startTick);
}

However, I receive this error:

error: must use '.*' or '->*' to call pointer-to-member function in '((__gnu_cxx::__ops::_Iter_pred<bool (MessageQueue::*)(const QueuedMessage&)>*)this)->__gnu_cxx::__ops::_Iter_pred<bool (MessageQueue::*)(const QueuedMessage&)>::_M_pred (...)', e.g. '(... ->* ((__gnu_cxx::__ops::_Iter_pred<bool (MessageQueue::*)(const QueuedMessage&)>*)this)->__gnu_cxx::__ops::_Iter_pred<bool (MessageQueue::*)(const QueuedMessage&)>::_M_pred) (...)'|

Is it a problem that I´m not using a vector of pointers? Or am I missing something? Thanks.

wasMessageSend is not a function. It's a member function. — Kerrek SB
– Kerrek SB, Commented Apr 23, 2017 at 15:12
wasMessageSend in the remove_if() call is a pointer to a member function. — user2100815
– user2100815, Commented Apr 23, 2017 at 15:13

songyuanyao · Accepted Answer · 2017-04-24 03:01:56Z

3

The predicate passed to std::remove_if is supposed to be a unary predicate with signature like bool pred(const Type &a);. That means it will be invoked with the element as the argument. But wasMessageSend is a member function so, besides the argument, it also needs an instance to be invoked on. So it can't be used with remove_if directly.

You can use lambda (since C++11) with it like:

_messages.erase(std::remove_if(
    _messages.begin(),
    _messages.end(),
    [this](const QueuedMessage& q) {
        return wasMessageSend(q);
    }
));

edited Apr 24, 2017 at 3:01

answered Apr 23, 2017 at 15:18

songyuanyao

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7 Comments

Pins Over a year ago

Thanks! It works! I think I totally forgot that this problem can be caused by using a member function as a comparator because I thought that it has a problem with a member function of the stored object not the class storing the vector.

songyuanyao Over a year ago

@BoundaryImposition Thanks for the editing! But why 3 spaces for indent?

Lightness Races in Orbit Over a year ago

@songyuanyao: Why not?

songyuanyao Over a year ago

@BoundaryImposition Because I'm just used to 2 or 4. Yes I know it's a personal opinion issue..

Lightness Races in Orbit Over a year ago

This is why tab-indentation is far superior, so that you can render the code anyway you choose without imposing your choice on other people (much like you can choose the font, and colour). Furthermore, the indentation of your code gains strict and unambiguous semantics (one tabspace = one level of indentation) that could be autonomously interpreted then handled in any way you like. Sadly, that works poorly on the web, so you are stuck with my three spaces. :)

|

A.S.H · Accepted Answer · 2017-04-23 15:31:01Z

1

Since wasMessageSend is a member function of your class MessageQueue, you need to "bind it" to an object (in this case, this) using std::bind

void MessageQueue::removeFromQueue(){
_messages.erase(std::remove_if(_messages.begin(),_messages.end(),
    std::bind(&MessageQueue::wasMessageSend, this, std::placeholders::_1), // bind this
    std::_messages.end());
}

Of course, this is not the only way, you may for example define a lambda predicate on the spot... but this method shows how to use a member function as a predicate.

edited Apr 23, 2017 at 15:31

answered Apr 23, 2017 at 15:20

A.S.H

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5 Comments

Lightness Races in Orbit Over a year ago

You may refer to song's answer

A.S.H Over a year ago

@BoundaryImposition I know that answer works too, but I'm eager to know why you consider it better or say "less antiquated" :). To my eyes this syntax is a bit simpler.

Lightness Races in Orbit Over a year ago

Looks a lot less simple to me. A lambda expression is very expressive, which is why it was added to the language

songyuanyao Over a year ago

@A.S.H You might want to read this post, or the book Effective Modern C++ Item 34: Prefer lambdas to std::bind for further infos.

A.S.H Over a year ago

I respect your opinions but I will always prefer to bind when the predicate is ready, there, but has the only "handicap" of being a member function. Thanks.

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