template functions?

Code:

---

#include <iostream>
#include <vector>
using namespace std;
template <class Type>
void print(const vector<Type>& list) {
  typedef vector<Type>::const_iterator Iter;
  for (Iter iter1 = list.begin(); iter1 != list.end(); ++iter1) {
      cout << *iter1 << " ";
  }
  cout << endl;
}

int main(void) {
  vector<int> nums;
  int i = 0;
  do {
      nums.push_back(i);
  } while (++i != 10);
  print(nums);
  return 0;
}

---

what I got from compiler message is test.cpp:
In function ‘void print(const std::vector<Type, std::allocator<_CharT> >&)’:
test.cpp:6: error: too few template-parameter-lists
test.cpp:7: error: ‘Iter’ was not declared in this scope
test.cpp:7: error: expected ‘;’ before ‘iter1’
test.cpp:7: error: ‘iter1’ was not declared in this scope
what can i resolve it?

I know it must be my misunderstanding about using template.. I just can't figure out why
ANy help will be really appreciated!

Try replacing:

Code:

---

  typedef vector<Type>::const_iterator Iter;

---

With:

Code:

---

  typedef typename vector<Type>::const_iterator Iter;

---

I'm not going in the details, in short, the compiler is too stupid (or, really, can't) to realize that "vector<Type>::const_iterator" is a type. Using "typename" the compiler is told that it is. That happens when you use template arguments like that.

It Works!! THX

One problem is that std::vector has two template parameters, not one. I am not absolutely sure about this, but it may even be possible for it to have more template parameters as a compiler extension. As such, it would be more typical to generalise this to take a range specified by iterators, and thus the template would be parameterised on the iterator type. Alternatively, you can use the container type as the template parameter, and thus the print function template will work for any container that has begin() and end() member functions that return a const_iterator.

Another problem is that the name vector<Type>::const_iterator does not necessarily refer to a type name. Since Type does not refer to a particular type, const_iterator could theoretically be a member name, and the compiler is supposed to assume as such, unless it is specified as a type name with the typename keyword:

Code:

---

typedef typename vector<Type>::const_iterator Iter;

---

Quote:

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Originally Posted by laserlight

One problem is that std::vector has two template parameters, not one. I am not absolutely sure about this, but it may even be possible for it to have more template parameters as a compiler extension. As such, it would be more typical to generalise this to take a range specified by iterators, and thus the template would be parameterised on the iterator type. Alternatively, you can use the container type as the template parameter, and thus the print function template will work for any container that has begin() and end() member functions that return a const_iterator.

---

That'd be no fun. Then he could just as well use:

Code:

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copy(nums.begin(), nums.end(), ostream_iterator<int>(cout, " "));

---

(not sure what include files are needed... iterator (for ostream_iterator) and algorithm (for copy), I think).

That basically does the same this function does (forgot to mention this earlier), except that it does not output a newline.

Quote:

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Originally Posted by EVOEx

That'd be no fun. Then he could just as well use:

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Not really. The name "print" arguably increases abstraction over the more general "copy", and then one removes the boilerplate concerning the use of ostream_iterator.

Quote:

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Originally Posted by EVOEx

(not sure what include files are needed... iterator (for ostream_iterator) and algorithm (for copy), I think).

---

Yes, and of course <iostream> for std::cout.

Quote:

---

Originally Posted by EVOEx

That basically does the same this function does (forgot to mention this earlier), except that it does not output a newline.

---

With my second option:

Code:

---

template<typename T>
void print(const T& list)
{
    using namespace std;
    typedef typename T::value_type value_type;
    copy(list.begin(), list.end(), ostream_iterator<value_type>(cout, " "));
    cout << endl;
}

// ...

vector<int> nums;
// ...
print(nums);

---

O_o I posted stuff, but now it is gone...

Why is it gone?!

Soma

Thx to both EVOEx and laserlight give such a good opinion.
Though I am not familiar with generic algorithm.. I'll learn it ASAP.
However,
I tried again as following..

Code:

---

50 template <class Type>
 51 void printResult(const Type& array) {
 52
 53    typedef typename Type::iterator Iter;
 54    //Print contents
 55    for (Iter iter1= array.begin(); iter1 != array.end(); ++iter1)
 56      cout << *iter1 << " ";
 57    cout << endl;

---

Code:

---

79 int main(void) {
 80    vector<int> numList;
 81    int i = 0;
 82    do {
 83      numList.push_back(i);
 84    } while (i++ != 10);
 85    printResult(numList);

---

And it didn't work..
test.cpp: In function ‘void printResult(const Type&) [with Type = std::vector<int, std::allocator<int> >]’:
test.cpp:85: instantiated from here
test.cpp:55: error: conversion from ‘__gnu_cxx::__normal_iterator<const int*, std::vector<int, std::allocator<int> > >’ to non-scalar type ‘printResult(const Type&) [with Type = std::vector<int, std::allocator<int> >]::Iter’ requested

Why didn't the compiler instantianted Type = vector<int> as I want?

The problem is probably because you are trying to use iterator when you should be using const_iterator.