output operator for vector

[Rares Vernica]

Hi,

I am trying to write a template that will take any type of vector and
output it in a certain way. Here is my code:

#include <ostream>
#include <vector>
#include <iterator>

using namespace std;

template <class T>
ostream& operator<<(ostr eam& out, const vector<Tv)
{
out << '[';
for(vector<T>:: const_iterator it = v.begin(); it != v.end(); it++) {
if (it != v.begin()) {
out << ", ";
}
out << *it;
}
out << ']';

return out;
}

I get the following error:

output.h: In function `std::ostream& operator<<(std: :ostream&,
std::vector<T, std::allocator< _CharT)':
output.h:22: error: expected `;' before "it"
output.h:22: error: `it' undeclared (first use this function)
output.h:22: error: (Each undeclared identifier is reported only once
for each function it appears in.)
make: *** [output.o]

Do you have any idea what the problem might me?

Thank you very much,
Ray

[Ian Collins]

Re: output operator for vector

Rares Vernica wrote:

Hi,
>
I am trying to write a template that will take any type of vector and
output it in a certain way. Here is my code:
>
#include <ostream>
#include <vector>
#include <iterator>
>
using namespace std;
>
template <class T>
ostream& operator<<(ostr eam& out, const vector<Tv)
{
out << '[';
for(vector<T>:: const_iterator it = v.begin(); it != v.end(); it++) {

should be

for(typename vector<T>::cons t_iterator it = v.begin(); it != v.end();
++it) {

Note I changed it++ to ++it, if an iterator isn't a simple pointer, ++it
is more efficient.


--
Ian Collins.

[Victor Bazarov]

Re: output operator for vector

Ian Collins wrote:

Rares Vernica wrote:

>Hi,
>>
>I am trying to write a template that will take any type of vector and
>output it in a certain way. Here is my code:
>>
>#include <ostream>
>#include <vector>
>#include <iterator>
>>
>using namespace std;
>>
>template <class T>
>ostream& operator<<(ostr eam& out, const vector<Tv)

It might also help to pass by a reference to const:

ostream& operator<<(ostr eam& out, const vector<T& v)

>{
> out << '[';
> for(vector<T>:: const_iterator it = v.begin(); it != v.end(); it++)
>{

>
should be
>
for(typename vector<T>::cons t_iterator it = v.begin(); it != v.end();
++it) {
>
Note I changed it++ to ++it, if an iterator isn't a simple pointer,
++it is more efficient.

The explanation for 'typename' is that 'const_iterator ' is a dependent
name. See FAQ.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[jhimcras@gmail.com]

Re: output operator for vector

When I compile that code in Visual C++ 6.0
There are no error.

[Rares Vernica]

Re: output operator for vector

Thank you very much,
Ray

Victor Bazarov wrote:

Ian Collins wrote:

>Rares Vernica wrote:

>>Hi,
>>>
>>I am trying to write a template that will take any type of vector and
>>output it in a certain way. Here is my code:
>>>
>>#include <ostream>
>>#include <vector>
>>#include <iterator>
>>>
>>using namespace std;
>>>
>>template <class T>
>>ostream& operator<<(ostr eam& out, const vector<Tv)

>
It might also help to pass by a reference to const:
>
ostream& operator<<(ostr eam& out, const vector<T& v)
>

>>{
>> out << '[';
>> for(vector<T>:: const_iterator it = v.begin(); it != v.end(); it++)
>>{

>should be
>>
>for(typename vector<T>::cons t_iterator it = v.begin(); it != v.end();
>++it) {
>>
>Note I changed it++ to ++it, if an iterator isn't a simple pointer,
>++it is more efficient.

>
The explanation for 'typename' is that 'const_iterator ' is a dependent
name. See FAQ.
>
V

[red floyd]

Re: output operator for vector

jhimcras@gmail. com wrote:

When I compile that code in Visual C++ 6.0
There are no error.
>

VC 6 is incorrect. As others have posted, a compliant compiler should
reject it without the typename keyword.