std::map iterators

[jason.cipriani]

In an std::map, iterators do not become invalid if other elements are
added erased. Just to make sure, does that mean that in the following
code:

map<string,B> m;
B *b = &(m["something"]);

The address in 'b' will remain valid, and always point to that same
object, as long as that data isn't erased from the map?

Similarily:

map<string,B> m;
map<string,B>::iterator i(m.find("something"));
B *b = (i == m.end()) ? NULL : &((*i).second);

Will b there also always remain valid as long as "something" remains
in the map?

Thanks,
Jason

 

[Frank Birbacher]

Hi!

The address in 'b' will remain valid, and always point to that same
object, as long as that data isn't erased from the map?
Yes.

Will b there also always remain valid as long as "something" remains
in the map?

Yes.

Regards,
Frank

 

[jason.cipriani]

[snip]

Thanks!

 

[Kai-Uwe Bux]

In an std::map, iterators do not become invalid if other elements are
added erased. Just to make sure, does that mean that in the following
code:

map<string,B> m;
B *b = &(m["something"]);

The address in 'b' will remain valid, and always point to that same
object, as long as that data isn't erased from the map?

Formally, validity of iterators and references into a map are separate
issues.

However, you are lucky: the standard also guarantees that references into a
map remain valid. Now, one could argue that reference validity does not
imply pointer validity. However, I think that is a little stretch. So 'b'
will remain valid. If you want to be on the formally safe side, you could
do something like:

B& b = m["something"];

Similarily:

map<string,B> m;
map<string,B>::iterator i(m.find("something"));
B *b = (i == m.end()) ? NULL : &((*i).second);

Will b there also always remain valid as long as "something" remains
in the map?

With the caveats from above: yes.


Best

Kai-Uwe Bux