reverse_iterato r and const_reverse_i terator

[Jess]

Hello,

I think the two reverse_iterato rs are the same, except that the const_
version doesn't allow me to change the value pointed to by the
iterators. However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

This one fails, with error

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

If I replace const_reverse_i terator with reverse_iterato r, then
everything works fine. Is there some subtle difference between the
two iterators?

Thanks,
Jess

[Roland Pibinger]

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

>I think the two reverse_iterato rs are the same, except that the const_
version doesn't allow me to change the value pointed to by the
iterators. However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

This one fails, with error

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

I guess your compiler cannot distinguish between the const and the
non-const rend(). Try:

for(vector<int> ::const_reverse _iterator i = v.rbegin();
i != ((const std::vector<int >) v).rend();
++i)
--
Roland Pibinger
"The best software is simple, elegant, and full of drama" - Grady Booch

[V. R. Marinov]

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

>If I replace const_reverse_i terator with reverse_iterato r, then
everything works fine. Is there some subtle difference between the
two iterators?

Yes. Basicly these are two different types/classes.

>However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

>for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

There are two overloaded rend(void) functions, that differ in their
return types. Now C++ does NOT permit function overloading by return
value. So the function "const_reverse_ iterator rend() const" could be
applied on const objects only, but not on non constant ones.
int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(); i!=v.rend(); ++i)
cout << (*i) << endl;
And the function "reverse_iterat or rend()" can be applied on non
constant objects only.

One way get around it is to explicitly cast the result returned by
rend(). And here is a different way deal with this:
int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(), End = v.end(); i!=End; ++i)
cout << (*i) << endl;

[Pete Becker]

Roland Pibinger wrote:

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

>I think the two reverse_iterato rs are the same, except that the const_
version doesn't allow me to change the value pointed to by the
iterators. However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

This one fails, with error

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

I guess your compiler cannot distinguish between the const and the
non-const rend(). Try:

Seems like it does distinguish between them, and calls the non-const
version on a non-const vector. The problem arises from comparing a
const_iterator to a plain iterator.

for(vector<int> ::const_reverse _iterator i = v.rbegin();
i != ((const std::vector<int >) v).rend();

i != ((const std::vector<int >&) v).rend();

That's a minor typo, but critical. As written, it copies the vector,
creating an iterator that points into the copy rather than the original.

A simpler way of doing this, if you have a library that conforms to the
not-yet-official C++0x standard, is to call crbegin() and crend()
instead of rbegin() and rend(). crbegin() and crend() return const
iterators.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)

[Jess]

On Jun 18, 4:17 am, Pete Becker <p...@versatile coding.comwrote :

Roland Pibinger wrote:

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

I think the two reverse_iterato rs are the same, except that the const_
version doesn't allow me to change the value pointed to by the
iterators. However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

This one fails, with error

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

I guess your compiler cannot distinguish between the const and the
non-const rend(). Try:

Seems like it does distinguish between them, and calls the non-const
version on a non-const vector. The problem arises from comparing a
const_iterator to a plain iterator.

for(vector<int> ::const_reverse _iterator i = v.rbegin();
i != ((const std::vector<int >) v).rend();

i != ((const std::vector<int >&) v).rend();

That's a minor typo, but critical. As written, it copies the vector,
creating an iterator that points into the copy rather than the original.

Thanks, I tried both methods, but neither worked... I think I only
need to cast a non-const "v" to a const "v", hence I tried

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i !=
(static_cast<co nst vector<int)v.re nd(); ++i)
cout << (*i) << endl;

Unfortunately, it failed as well...Moreover , what's the difference
between casting to a class type and casting to a reference?

Jess

[Jess]

On Jun 18, 4:10 am, "V. R. Marinov" <v.r.mari...@gm ail.comwrote:

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

>If I replace const_reverse_i terator with reverse_iterato r, then
>everything works fine. Is there some subtle difference between the
>two iterators?

Yes. Basicly these are two different types/classes.

>However, I have a program that works for reverse_iterato r
>but not const_reverse_i terator:

>for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
>++i)
cout << (*i) << endl;

There are two overloaded rend(void) functions, that differ in their
return types. Now C++ does NOT permit function overloading by return
value. So the function "const_reverse_ iterator rend() const" could be
applied on const objects only, but not on non constant ones.

int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(); i!=v.rend(); ++i)
cout << (*i) << endl;

And the function "reverse_iterat or rend()" can be applied on non
constant objects only.

One way get around it is to explicitly cast the result returned by
rend(). And here is a different way deal with this:

int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(), End = v.end(); i!=End; ++i)
cout << (*i) << endl;

Thanks. By "explicitly cast the result returned by rend()", do you
mean I should cast const away? Is it done by using "v.end()" instead
of "rend()"? However, why do we need to cast away constness if "v" is
a const vector?

Jess

[V. R. Marinov]

Jess напи�а:

On Jun 18, 4:10 am, "V. R. Marinov" <v.r.mari...@gm ail.comwrote:

>On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:

> >If I replace const_reverse_i terator with reverse_iterato r, then
everything works fine. Is there some subtle difference between the
two iterators?

Yes. Basicly these are two different types/classes.

> >However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

> >for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

There are two overloaded rend(void) functions, that differ in their
return types. Now C++ does NOT permit function overloading by return
value. So the function "const_reverse_ iterator rend() const" could be
applied on const objects only, but not on non constant ones.

int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(); i!=v.rend(); ++i)
cout << (*i) << endl;

And the function "reverse_iterat or rend()" can be applied on non
constant objects only.

One way get around it is to explicitly cast the result returned by
rend(). And here is a different way deal with this:

int arr[] = {1, 2, 3};
const vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;
for(ViCRI i = v.rbegin(), End = v.end(); i!=End; ++i)
cout << (*i) << endl;

Thanks. By "explicitly cast the result returned by rend()", do you
mean I should cast const away? Is it done by using "v.end()" instead
of "rend()"? However, why do we need to cast away constness if "v" is
a const vector?

No. I was referring to the non-const case.

/*************** ******* CODE *************** *******/
int arr[] = {1, 2, 3};
vector<intv(arr , arr+3);

typedef vector<int>::co nst_reverse_ite rator ViCRI;

for(ViCRI i = v.rbegin(); i!=ViCRI(v.rend ()); ++i)
cout << (*i) << endl;
/*************** ******* CODE *************** *******/

[V.R. Marinov]

Jess :

On Jun 18, 4:17 am, Pete Becker <p...@versatile coding.comwrote :

Roland Pibinger wrote:

On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:
>I think the two reverse_iterato rs are the same, except that the const_
>version doesn't allow me to change the value pointed to by the
>iterators. However, I have a program that works for reverse_iterato r
>but not const_reverse_i terator:

>for(vector<int >::const_revers e_iterator i = v.rbegin(); i != v.rend();
>++i)
> cout << (*i) << endl;

>This one fails, with error

>no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
>[with _Tp = int, _Alloc = std::allocator< int>]()'

I guess your compiler cannot distinguish between the const and the
non-const rend(). Try:

Seems like it does distinguish between them, and calls the non-const
version on a non-const vector. The problem arises from comparing a
const_iterator to a plain iterator.

for(vector<int> ::const_reverse _iterator i = v.rbegin();
i != ((const std::vector<int >) v).rend();

i != ((const std::vector<int >&) v).rend();

That's a minor typo, but critical. As written, it copies the vector,
creating an iterator that points into the copy rather than the original.

Thanks, I tried both methods, but neither worked... I think I only
need to cast a non-const "v" to a const "v", hence I tried

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i !=
(static_cast<co nst vector<int)v.re nd(); ++i)
cout << (*i) << endl;

Unfortunately, it failed as well

It looks like you are using the wrong syntax. It should be.

i != static_cast<con st vector<int>& >(v).rend()

>...Moreover, what's the difference
between casting to a class type and casting to a reference?

Please reread Pete Becker's post.

[Roland Pibinger]

On Sun, 17 Jun 2007 15:48:43 -0700, Jess wrote:

>Thanks, I tried both methods, but neither worked... I think I only
need to cast a non-const "v" to a const "v", hence I tried

for(vector<int >::const_revers e_iterator i = v.rbegin(); i !=
(static_cast<c onst vector<int)v.re nd(); ++i)
cout << (*i) << endl;

Actually, the cast is a bad idea anyway (I used it only to find the
cause of the problem). Better use a reference to const vector on which
only const member functions can be invoked:

vector<intv;
// ...
const vector<int>& cv = v; // reference
for(vector<int> ::const_reverse _iterator i = cv.rbegin();
i != cv.rend(); ++i)
cout << (*i) << endl;

--
Roland Pibinger
"The best software is simple, elegant, and full of drama" - Grady Booch

[Jess]

On Jun 18, 2:35 pm, "V.R. Marinov" <v.r.mari...@gm ail.comwrote:

Jess :

On Jun 18, 4:17 am, Pete Becker <p...@versatile coding.comwrote :

Roland Pibinger wrote:
On Sun, 17 Jun 2007 06:45:56 -0700, Jess wrote:
I think the two reverse_iterato rs are the same, except that the const_
version doesn't allow me to change the value pointed to by the
iterators. However, I have a program that works for reverse_iterato r
but not const_reverse_i terator:

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i != v.rend();
++i)
cout << (*i) << endl;

This one fails, with error

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

I guess your compiler cannot distinguish between the const and the
non-const rend(). Try:

Seems like it does distinguish between them, and calls the non-const
version on a non-const vector. The problem arises from comparing a
const_iterator to a plain iterator.

for(vector<int> ::const_reverse _iterator i = v.rbegin();
i != ((const std::vector<int >) v).rend();

i != ((const std::vector<int >&) v).rend();

That's a minor typo, but critical. As written, it copies the vector,
creating an iterator that points into the copy rather than the original.

Thanks, I tried both methods, but neither worked... I think I only
need to cast a non-const "v" to a const "v", hence I tried

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i !=
(static_cast<co nst vector<int)v.re nd(); ++i)
cout << (*i) << endl;

Unfortunately, it failed as well

It looks like you are using the wrong syntax. It should be.

i != static_cast<con st vector<int>& >(v).rend()

Thanks. I tried but it still doesn't work. Here's the entire
program.

#include<vector >
#include<string >
#include<iterat or>
#include<iostre am>
using namespace std;

int main(){
vector<intv;
for(int i = 0; i != 10; ++i)
v.push_back(i);

for(vector<int> ::const_reverse _iterator i = v.rbegin(); i !=
static_cast<vec tor<int>& >(v).rend(); ++i)
cout << (*i) << endl;
return 0;
}

The error message was

no match for 'operator!=' in 'i != std::vector<_Tp , _Alloc>::rend()
[with _Tp = int, _Alloc = std::allocator< int>]()'

Jess