template error with iterator

[utab]

Dear all,

I was experimenting with a template taking two iterators for the range
of
a vector.(Perhaps , it is sth simple and I am missing it because it is
a
late hour.) I ran into problems in the compile phase , the code is
below:

#include <iostream>
#include <algorithm>
#include <stdexcept>
#include <vector>

using std::domain_err or;
using std::sort;
using std::vector;

template <class T, class Ran>
T median(Ran b, Ran e)
{
typedef typename vector<T>::size _type vec_sz;

vec_sz size = (e-b)/sizeof(T);
if (size == 0)
throw domain_error("m edian of an empty vector");

sort(b, e);

vec_sz mid = size/2;

return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
}

int main()
{
vector<doubleve c;
for(int i=0;i!=10;++i)
vec.push_back(i );
std::cout << median(vec.begi n(), vec.end()) << std::endl;
return 0;
}


I get

83.cc: In function 'int main()':
83.cc:38: error: no matching function for call to
'median(__gnu_c xx::__normal_it erator<double*, std::vector<dou ble,
std::allocator< double >, __gnu_cxx::__no rmal_iterator<d ouble*,
std::vector<dou ble, std::allocator< double >)'

I am supplying the median function with iterators by using the begin
and
end member functions of the vector. However, I get a type mismatch
error
on iterators I guess. Could you clarify the problem for me?

Rgds,

--
Umut

[utab]

Re: template error with iterator

On May 13, 1:37 am, utab <umut.ta...@gma il.comwrote:

Dear all,
>
I was experimenting with a template taking two iterators for the range
of
a vector.(Perhaps , it is sth simple and I am missing it because it is
a
late hour.) I ran into problems in the compile phase , the code is
below:
>
#include <iostream>
#include <algorithm>
#include <stdexcept>
#include <vector>
>
using std::domain_err or;
using std::sort;
using std::vector;
>
template <class T, class Ran>
T median(Ran b, Ran e)
{
typedef typename vector<T>::size _type vec_sz;
>
vec_sz size = (e-b)/sizeof(T);
if (size == 0)
throw domain_error("m edian of an empty vector");
>
sort(b, e);
>
vec_sz mid = size/2;
>
return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
>
}
>
int main()
{
vector<doubleve c;
for(int i=0;i!=10;++i)
vec.push_back(i );
std::cout << median(vec.begi n(), vec.end()) << std::endl;
return 0;
>
}
>
I get
>
83.cc: In function 'int main()':
83.cc:38: error: no matching function for call to
'median(__gnu_c xx::__normal_it erator<double*, std::vector<dou ble,
std::allocator< double >, __gnu_cxx::__no rmal_iterator<d ouble*,
std::vector<dou ble, std::allocator< double >)'
>
I am supplying the median function with iterators by using the begin
and
end member functions of the vector. However, I get a type mismatch
error
on iterators I guess. Could you clarify the problem for me?
>
Rgds,
>
--
Umut

Forgot to tell that I use g++ 4.1.2

[tragomaskhalos]

Re: template error with iterator

On 13 May, 00:37, utab <umut.ta...@gma il.comwrote:

>
template <class T, class Ran>
T median(Ran b, Ran e)
{
  typedef typename vector<T>::size _type vec_sz;
>
  vec_sz size = (e-b)/sizeof(T);
  if (size == 0)
    throw domain_error("m edian of an empty vector");
>
  sort(b, e);
>
  vec_sz mid = size/2;
>
  return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
>
}
>
int main()
{
  vector<doubleve c;
  for(int i=0;i!=10;++i)
    vec.push_back(i );
  std::cout << median(vec.begi n(), vec.end()) << std::endl;

The compiler cannot deduce T; replace the above call with
median<double>( vec.begin(), vec.end())
^^^^^^

  return 0;
>
}
>

[utab]

Re: template error with iterator

On May 13, 1:47 am, tragomaskhalos <dave.du.verg.. .@logicacmg.com >
wrote:

On 13 May, 00:37, utab <umut.ta...@gma il.comwrote:
>
>
>
>
>

template <class T, class Ran>
T median(Ran b, Ran e)
{
typedef typename vector<T>::size _type vec_sz;

>

vec_sz size = (e-b)/sizeof(T);
if (size == 0)
throw domain_error("m edian of an empty vector");

>

sort(b, e);

>

vec_sz mid = size/2;

>

return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];

>

}

>

int main()
{
vector<doubleve c;
for(int i=0;i!=10;++i)
vec.push_back(i );
std::cout << median(vec.begi n(), vec.end()) << std::endl;

>
The compiler cannot deduce T; replace the above call with
median<double>( vec.begin(), vec.end())
^^^^^^
>

return 0;

>

}

Thanks, I read something similar in the faq as well.

[acehreli@gmail.com]

Re: template error with iterator

On May 12, 4:37 pm, utab <umut.ta...@gma il.comwrote:

template <class T, class Ran>
T median(Ran b, Ran e)
{

According to that definition, there is no relation between T and Ran.

[...]

std::cout << median(vec.begi n(), vec.end()) << std::endl;

The complier cannot deduce T for that call because the return values
are never a part of function signatures. You can try specifying it
yourself:

median<double>( vec.begin(), vec.end());

Ali

[Paavo Helde]

Re: template error with iterator

utab <umut.tabak@gma il.comwrote in news:50129083-18fd-4099-9ba7-
586bb9df72e2@a2 3g2000hsc.googl egroups.com:

Dear all,
>
I was experimenting with a template taking two iterators for the range
of
a vector.(Perhaps , it is sth simple and I am missing it because it is
a
late hour.) I ran into problems in the compile phase , the code is
below:
>
#include <iostream>
#include <algorithm>
#include <stdexcept>
#include <vector>
>
using std::domain_err or;
using std::sort;
using std::vector;
>
template <class T, class Ran>
T median(Ran b, Ran e)
{

You are passing in only Ran arguments, there is no way the compiler could
deduce the T template parameter.

typedef typename vector<T>::size _type vec_sz;
>
vec_sz size = (e-b)/sizeof(T);
if (size == 0)
throw domain_error("m edian of an empty vector");
>
sort(b, e);
>
vec_sz mid = size/2;
>
return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
}
>
int main()
{
vector<doubleve c;
for(int i=0;i!=10;++i)
vec.push_back(i );
std::cout << median(vec.begi n(), vec.end()) << std::endl;

... so you have to specify it yourself:

std::cout << median<double>( vec.begin(), vec.end()) << std::endl;

... or throw away the redundant template parameter:

template <class Ran>
typename Ran::value_type median(Ran b, Ran e)
{
typedef typename vector<Ran::val ue_type>::size_ type vec_sz;

vec_sz size = (e-b)/sizeof(Ran::val ue_type);
if (size == 0)
throw domain_error("m edian of an empty vector");

....

hth
Paavo

[Greg Herlihy]

Re: template error with iterator

On May 12, 4:37 pm, utab <umut.ta...@gma il.comwrote:

I was experimenting with a template taking two iterators for the range
of
a vector.(Perhaps , it is sth simple and I am missing it because it is
a
late hour.) I ran into problems in the compile phase , the code is
below:
>
#include <iostream>
#include <algorithm>
#include <stdexcept>
#include <vector>
>
using std::domain_err or;
using std::sort;
using std::vector;
>
template <class T, class Ran>
T median(Ran b, Ran e)
{
  typedef typename vector<T>::size _type vec_sz;
>
  vec_sz size = (e-b)/sizeof(T);
  if (size == 0)
    throw domain_error("m edian of an empty vector");
>
  sort(b, e);
>
  vec_sz mid = size/2;
>
  return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
}

The C++ compiler is not able to deduce the type "T" from the median()
function call. Since "T" is dependent on the iterator type "Ran", I
would eliminate "T" altogether. Note also that the calculation of
"size" is incorrect - e-b will return the number of positions between
two random access iterators.

Applying these suggestions produces a program like this one:

#include <iostream>
#include <algorithm>
#include <stdexcept>
#include <vector>
#include <iterator>

using std::domain_err or;
using std::sort;
using std::vector;

template <class Ran>
typename std::iterator_t raits<Ran>::val ue_type
median(Ran b, Ran e)
{
size_t size = e-b;
if (size == 0)
throw domain_error("m edian of an empty vector");
sort(b, e);
size_t mid = size/2;
return size % 2 == 0 ? (b[mid] + b[mid-1]) / 2 : b[mid];
}

int main()
{
vector<doubleve c;

for (int i=0; i!=10; ++i)
vec.push_back(i );

std::cout << median(vec.begi n(), vec.end()) << std::endl;
return 0;
}

Greg

[James Kanze]

Re: template error with iterator

On May 13, 1:57 am, Paavo Helde <nob...@ebi.eew rote:

utab <umut.ta...@gma il.comwrote in news:50129083-18fd-4099-9ba7-
586bb9df7...@a2 3g2000hsc.googl egroups.com:

I was experimenting with a template taking two iterators for
the range of a vector.(Perhaps , it is sth simple and I am
missing it because it is a late hour.) I ran into problems
in the compile phase , the code is below:

[...]

.. or throw away the redundant template parameter:

template <class Ran>
typename Ran::value_type median(Ran b, Ran e)

There's no guarantee that Ran has a typedef for value_type (and
there have definitely been implementations of the standard
library where it didn't for std::vector<>).

This should be:

template< typename RandomAccessIte rator >
typename std::iterator_t raits< RandomAccessIte rator >::value_type
median( RandomAccessIte rator begin, RandomAccessIte rator end )

{
typedef typename vector<Ran::val ue_type>::size_ type vec_sz;

And here, even worse, since you're tying yourself to
std::vector. (Even as it stands, you need an additional
typename in the template argument.) Also, the difference
between to random access iterators is a difference_type , not a
size_type. So it should be:

typedef typename std::iterator_t raits< RandomAccessIte rator >
::difference_ty pe SequenceSize ;

vec_sz size = (e-b)/sizeof(Ran::val ue_type);

And of course, you don't want the division.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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