std::map

[Duane Hebert]

2 questions:

Given a map defined as

std::map<int,string> stringmap;

//How do you access the data_type (string) via an iterator?

std::string spoo("doh");

stringmap.insert(std::make_pair(1,spoo));
// I tried this:
for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;
But this doesn't compile with error that the operator+ has not been defined
....
I think the problem is that the iterator returns a pair. Is it possible to
do this?

I know I can do:
if(stringmap.find(1) != stringmap.end()) std::cout << stringmap[1];
2nd question:

This doesn't work with BCB6 (causes a compile error about casting a pair to
an int???)

//stringmap is defined as a private class member
const std::string classname::getstring(size_t index) const { return
stringmap[index];}

but this compiles fine and seems to work with no problems:

const std::string classname::getstring(size_t index) { return
stringmap[index];}

Is this correct? Why can I not define this function as const? It's not
changing any
member data.

Thanks.

[Christoph Rabel]

Duane Hebert wrote:

// I tried this:
for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;
std::cout << it->second;

should do the trick.

Btw I recommend that you use a const_iterator instead of an
iterator.

I know I can do:
if(stringmap.find(1) != stringmap.end()) std::cout << stringmap[1];
Brr, this is terribly inefficient. Why dont you store the
returnvalue of find?

if((it = stringmap.find(1)) != stringmap.end())
std::cout << it->second;

2nd question:

This doesn't work with BCB6 (causes a compile error about casting a pair to
an int???)

//stringmap is defined as a private class member
const std::string classname::getstring(size_t index) const { return
stringmap[index];}

but this compiles fine and seems to work with no problems:

const std::string classname::getstring(size_t index) { return
stringmap[index];}

Is this correct? Why can I not define this function as const? It's not
changing any member data.

Because operator[] is not const. It might change the
contents of your stringmap. The map is const in this
function, so you are not allowed to call it.

Use find instead.

hope this helps,

Christoph

[Rob Williscroft]

Duane Hebert wrote in news:YY******************@wagner.videotron.net:

2 questions:

Given a map defined as

std::map<int,string> stringmap;

//How do you access the data_type (string) via an iterator?

std::string spoo("doh");

stringmap.insert(std::make_pair(1,spoo));
// I tried this:
for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;
Change this to it->second; if you want the key do it->first.


But this doesn't compile with error that the operator+ has not been
defined ...
I think the problem is that the iterator returns a pair. Is it
possible to do this?

I know I can do:
if(stringmap.find(1) != stringmap.end()) std::cout <<
stringmap[1];


This should work fine stringmap[i] should return a std::string &
2nd question:

This doesn't work with BCB6 (causes a compile error about casting a
pair to an int???)
I don't really understand this, maybe you should post a complete
example that doesen't compile.

//stringmap is defined as a private class member
const std::string classname::getstring(size_t index) const { return
stringmap[index];}

A return of std::string const & would be more effecient here.

but this compiles fine and seems to work with no problems:

const std::string classname::getstring(size_t index) { return
stringmap[index];}

Is this correct? Why can I not define this function as const? It's
not changing any
member data.

Yes this is correct std::map<>::operator[] is not defined as a
const-member function, this is because the non-const version
will insert a new element into the map if it doesn't exist.

std::string classname::getstring( int index ) const
{
std::map<int,string>::iterator ptr =
stringmap.find( index )
:

if ( ptr != stringmap.end() ) return ptr->second;

// return std::string( "< some default>" );
//throw std::runtime_error( "Item Not Found" );

// the choice is your's.
}
HTH

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Smeckler]

> > const std::string classname::getstring(size_t index) { return

stringmap[index];}

Is this correct? Why can I not define this function as const? It's not
changing any member data.

Because operator[] is not const. It might change the
contents of your stringmap. The map is const in this
function, so you are not allowed to call it.

But why doesn't map define a const operator[] in addition to the non const
one?

[Smeckler]

> > But why doesn't map define a const operator[] in addition to the non
const

one?

What would it do if the key was not present in the map?

Errrr...do nothing and return a reference to NULL :)

[jeffc]

"Duane Hebert" <sp**@flarn.com> wrote in message
news:YY******************@wagner.videotron.net...

2 questions:

Given a map defined as

std::map<int,string> stringmap;

//How do you access the data_type (string) via an iterator?

std::string spoo("doh");

stringmap.insert(std::make_pair(1,spoo));
// I tried this:
for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;

So you've figured out that you need to insert pairs, so the thing in the map
that the iterator is pointing to is a pair. You also said you want only the
string, not the whole pair. *it would be the whole pair. In a pair, there
is a "first" and "second". You want to point to the "second".

[Chris Theis]

"Smeckler" <sm*********@hotSPAMmail.com> wrote in message
news:bj*******************@news.demon.co.uk...

But why doesn't map define a const operator[] in addition to the non const one?

What would it do if the key was not present in the map?

Errrr...do nothing and return a reference to NULL :)

And how do you return a _reference_ to NULL?

Chris

[Ron Natalie]

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message news:bj**********@sunnews.cern.ch...

"Smeckler" <sm*********@hotSPAMmail.com> wrote in message
news:bj*******************@news.demon.co.uk...

> But why doesn't map define a const operator[] in addition to the non

const

> one?
>
What would it do if the key was not present in the map?

Errrr...do nothing and return a reference to NULL :)

And how do you return a _reference_ to NULL?

That was a joke, son. (Didn't you see the smilely?)

[Duane Hebert]

> > // I tried this:

for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;
Change this to it->second; if you want the key do it->first.

Thanks. I stumbled on that myself. I also found that
std::cout (*it).second; works as well. This was what I tried
originally but I missed the () .

I know I can do:
if(stringmap.find(1) != stringmap.end()) std::cout <<
stringmap[1];

This should work fine stringmap[i] should return a std::string &

That's correct but if I want to access each element of the map with a for
loop
it requires that I know the min and max keys and try all of them. That's
why
I wanted to get the iterator to work. (also to figure it out :-)

//stringmap is defined as a private class member
const std::string classname::getstring(size_t index) const { return
stringmap[index];}

A return of std::string const & would be more effecient here.

but this compiles fine and seems to work with no problems:

const std::string classname::getstring(size_t index) { return
stringmap[index];}

Is this correct? Why can I not define this function as const? It's
not changing any
member data.

Yes this is correct std::map<>::operator[] is not defined as a
const-member function, this is because the non-const version
will insert a new element into the map if it doesn't exist.

std::string classname::getstring( int index ) const
{
std::map<int,string>::iterator ptr =
stringmap.find( index )
:

if ( ptr != stringmap.end() ) return ptr->second;

This didn't exactly work on my compiler. Threw the same error as before.
I had to make ptr a const_iterator,
then it worked fine. Not sure why since find() returns an iterator,
but thanks for the help.
HTH

It does. Thanks.

[Christoph Rabel]

Duane Hebert wrote:

std::string classname::getstring( int index ) const
{
std::map<int,string>::iterator ptr =
stringmap.find( index )
:

if ( ptr != stringmap.end() ) return ptr->second;

This didn't exactly work on my compiler. Threw the same error as before.
I had to make ptr a const_iterator,
then it worked fine. Not sure why since find() returns an iterator,
but thanks for the help.

No, because the stringmap is const in this function find
returns a const_iterator in this case.

Christoph

[Rob Williscroft]

Duane Hebert wrote in news:Ii*******************@wagner.videotron.net:

std::string classname::getstring( int index ) const

{
std::map<int,string>::iterator ptr =
stringmap.find( index )
:

if ( ptr != stringmap.end() ) return ptr->second;

This didn't exactly work on my compiler. Threw the same error as
before. I had to make ptr a const_iterator,
then it worked fine. Not sure why since find() returns an iterator,
but thanks for the help.

My code was wrong, but fortunatly your compiler cought the error.

Since this is a const member function the member stingmap is
effectivly constant within it so stringmap.find( index ) is calling
std::map<>::find( Key ) const, which returns a const_iterator.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Duane Hebert]

> My code was wrong, but fortunatly your compiler cought the error.

Since this is a const member function the member stingmap is
effectivly constant within it so stringmap.find( index ) is calling
std::map<>::find( Key ) const, which returns a const_iterator.

Borland 's BCB6 uses StlPort which seems to be pretty compliant and
less buggy than the RogueWave implementations of previous editions.
The help for the stl containers is not the best though. I'll be getting a
couple of books soon.

[Duane Hebert]

> > // I tried this:

for(std::map<int,string> ::iterator it = stringmap.begin(); it !=
stringmap.end(); ++it)
std::cout << *it;
So you've figured out that you need to insert pairs, so the thing in the

map that the iterator is pointing to is a pair. You also said you want only the string, not the whole pair. *it would be the whole pair. In a pair, there is a "first" and "second". You want to point to the "second".

Right but *it.second doesn't work. (second is not a member of iterator)
This was my problem. I could create a
pair from *it and it would work. I didn't realize that there are two
operators
in the line *it.second, the . and the * and the . has precedence. So this
line
turns into *(it.second).
One solution would be to do it->first since it returns a reference.
I don't like this because I think that it obscures the fact that *it is a
reference.
The solution that I was looking for was simply (*it).second.

The answers that I received here were very helpful. std::map is different
from containers
like vectors and sets in that it contains another "container".

[Kevin Goodsell]

Duane Hebert wrote:

Duane, look at the formatting of your messages. Are you hitting <enter>
at the end of your lines or something? Please don't do that. Set your
news client's line wrap to a reasonable length (around 72 is good) and
let it do the wrapping.

Right but *it.second doesn't work. (second is not a member of iterator)
This was my problem. I could create a
pair from *it and it would work. I didn't realize that there are two
operators
in the line *it.second, the . and the * and the . has precedence. So this
line
turns into *(it.second).
One solution would be to do it->first since it returns a reference.
I don't like this because I think that it obscures the fact that *it is a
reference.

It absolutely does not. Every C & C++ programmer recognizes immediately
that a->b means (*a).b. If anything the 'arrow' syntax is less confusing.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Smeckler]

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message
news:bj**********@sunnews.cern.ch...

"Smeckler" <sm*********@hotSPAMmail.com> wrote in message
news:bj*******************@news.demon.co.uk...

> But why doesn't map define a const operator[] in addition to the non

const

> one?
>
What would it do if the key was not present in the map?

Errrr...do nothing and return a reference to NULL :)

And how do you return a _reference_ to NULL?

Chris

Easy. And, since this is the STL we're talking about, why not create a
handy template function :)

template <typename TypeName> TypeName& GetNullReference()
{
return *(TypeName*)NULL;
}

[jeffc]

"Duane Hebert" <sp**@flarn.com> wrote in message
news:ns******************@weber.videotron.net...

So you've figured out that you need to insert pairs, so the thing in the map

that the iterator is pointing to is a pair. You also said you want only

the

string, not the whole pair. *it would be the whole pair. In a pair,

there

is a "first" and "second". You want to point to the "second".

Right but *it.second doesn't work.

So?
One solution would be to do it->first since it returns a reference.
I don't like this because I think that it obscures the fact that *it is a
reference.
The solution that I was looking for was simply (*it).second.

Yuck. Why would you want to muck it up like that? it->second is perfectly
natural. I think it's crazy to say pointer notation "obscures" the fact
that the dereferenced pointer is a reference. That's like saying you prefer
the statement "It's false that I don't like pie" to the statement "I like
pie." because the second sentence obscures the fact that you don't dislike
it.

[llewelly]

"Smeckler" <sm*********@hotSPAMmail.com> writes:
[snip]

Easy. And, since this is the STL we're talking about, why not create a
handy template function :)

template <typename TypeName> TypeName& GetNullReference()
{
return *(TypeName*)NULL;
}

This has undefined behavior.

[Kevin Goodsell]

Smeckler wrote:

Easy. And, since this is the STL we're talking about, why not create a
handy template function :)

template <typename TypeName> TypeName& GetNullReference()
{
return *(TypeName*)NULL;
}

I hope you are joking. If not, you need to learn some C++.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Duane Hebert]

> Yuck. Why would you want to muck it up like that? it->second is
perfectly

natural. I think it's crazy to say pointer notation "obscures" the fact
that the dereferenced pointer is a reference. That's like saying you prefer the statement "It's false that I don't like pie" to the statement "I like
pie." because the second sentence obscures the fact that you don't dislike
it.

I prefer to use the dot operator when referencing an object that's not a
pointer. It's
a matter of taste and not worth an argument. Both ways work. To me, it
wasn't intuitive
that (*it).first would work. But to answer your simile, I guess that it's
not true that I do
agree with you :-)

[Smeckler]

> > Easy. And, since this is the STL we're talking about, why not create a

handy template function :)

template <typename TypeName> TypeName& GetNullReference()
{
return *(TypeName*)NULL;
}

I hope you are joking. If not, you need to learn some C++.

-Kevin

Kevin,

Would you care to be more specific about your objection to my code?

[Christoph Rabel]

Smeckler wrote:

Would you care to be more specific about your objection to my code?

First, this has undefined behaviour. (Dereferencing a 0 Pointer)

Second, how do you test the returned reference for 0??
You cannot. So somebody ends up with a dangling reference. And probably
crashes a while later when he tries to use it...

Christoph

[tom_usenet]

On Tue, 9 Sep 2003 10:20:27 +0100, "Smeckler"
<sm*********@hotSPAMmail.com> wrote:

> Easy. And, since this is the STL we're talking about, why not create a
> handy template function :)
>
> template <typename TypeName> TypeName& GetNullReference()
> {
> return *(TypeName*)NULL;
> }
>
>

I hope you are joking. If not, you need to learn some C++.

-Kevin

Kevin,

Would you care to be more specific about your objection to my code?

It has undefined behaviour. Dereferencing a null pointer is always a
bug. In other words, in the language spec, there is no such thing as a
null reference, since in order to create one, you have to go outside
the language spec.

In practice, the code will crash on some platforms, while on others it
will return what you think of as a "null reference".

Tom

[jeffc]

"Duane Hebert" <sp**@flarn.com> wrote in message
news:I_*******************@weber.videotron.net...

I prefer to use the dot operator when referencing an object that's not a
pointer. It's
a matter of taste and not worth an argument. Both ways work. To me, it
wasn't intuitive
that (*it).first would work.

The whole point is it's supposed to look like a pointer. Otherwise, you
wouldn't have to go around your elbow to get to your butt to dereference it,
thereby forcing it to look like a reference just because that's how you want
it to look.

[jeffc]

"Duane Hebert" <sp**@flarn.com> wrote in message
news:I_*******************@weber.videotron.net...

I prefer to use the dot operator when referencing an object that's not a
pointer. It's
a matter of taste and not worth an argument. Both ways work. To me, it
wasn't intuitive
that (*it).first would work.

Or to put it another way, if it were just a regular reference object, you
would access it like
it.
instead of like
(*it).
So it's not just a matter of preference in this case, it's a matte of how
the thing was designed to be used - i.e., like a pointer.

[Kevin Goodsell]

Smeckler wrote:

Easy. And, since this is the STL we're talking about, why not create a
handy template function :)

template <typename TypeName> TypeName& GetNullReference()
{
return *(TypeName*)NULL;
}

I hope you are joking. If not, you need to learn some C++.

-Kevin

Kevin,

Would you care to be more specific about your objection to my code?

I could, but I think I'll let Herb Sutter do it instead:

http://www.gotw.ca/conv/002.htm

To summarize: You cannot have a null reference. You cannot dereference a
null pointer, under any circumstances.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.