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<<(ostream& 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]

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

should be

for(typename vector<T>::const_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]

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<<(ostream& out, const vector<Tv)

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

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

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

should be

for(typename vector<T>::const_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
--
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[jhimcras]

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

[Rares Vernica]

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<<(ostream& out, const vector<Tv)

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

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

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

should be

for(typename vector<T>::const_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]

jh******@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.