set<int> having FIFO behaviour

[Luca]

I have the need of a container of integers showing both the
characteristics of an associative container (all integer elements
different from each other) and the FIFO behaviour.

Do you know if there is a ready-made container for that or if I have
to use - for example - a set<int> and produce by myself FIFO
behaviour, or maybe I can use a queue<int> adding code for preventing
the insertion od duplicate integers

Thank you

Luca

 

[Ron Natalie]

I have the need of a container of integers showing both the
characteristics of an associative container (all integer elements
different from each other) and the FIFO behaviour.

Do you know if there is a ready-made container for that or if I have
to use - for example - a set<int> and produce by myself FIFO
behaviour, or maybe I can use a queue<int> adding code for preventing
the insertion od duplicate integers

You can't order a set other than the way implied by the comparison function
it is created with (usually Less). The easiest way is what you suggest
in the second paragraph. One easy way to check for duplication is to
use both the queue and set. Check the set prioir to insert, insert into
both if not found. Delete from the set when you delete from the queue.

 

[Jerry Coffin]

I have the need of a container of integers showing both the
characteristics of an associative container (all integer elements
different from each other) and the FIFO behaviour.

It sounds like a priority queue should work for you.

 

[Koenings Bernhard]

Hi Luca!

I have the need of a container of integers showing both the
characteristics of an associative container (all integer elements
different from each other) and the FIFO behaviour.

Do you know if there is a ready-made container for that or if I have
to use - for example - a set<int> and produce by myself FIFO
behaviour, or maybe I can use a queue<int> adding code for preventing
the insertion od duplicate integers

Try using list as container. But there's a problem with the uniqueness. You
have to call the 'unique' method after each insert, that may result in bad
performance

HTH
Bernhard

 

[Kevin Goodsell]

Try using list as container. But there's a problem with the uniqueness. You
have to call the 'unique' method after each insert, that may result in bad
performance

I don't think that will work. 'unique' can only find duplicates that are
adjacent, implying that you must usually sort the container first. This
would destroy the existing order.

-Kevin