CODEWITHSUNDEEP
c++iterator
Enamul Hassan
Enamul Hassan

Reputation: 5445

how the iterator in c++ could be printed?

Suppose, I have declared a vector in C++ like this:

vector<int>numbers = {4,5,3,2,5,42};

I can iterate it through the following code:

for (vector<int>::iterator it = numbers.begin(); it!=numbers.end(); it++){
    // code goes here
}

Now, I would talk about coding in the block of for loop.

I can access and change any value using this iterator. say, I want to increase every value by 10 and the print. So, the code would be:

*it+=10;
cout << *it << endl;

I can print the address of both iterator and elements that are being iterated.

Address of iterator can be printed by:

cout << &it << endl;

Address of iterated elements can be printed by:

cout << &(*it) << endl;

But why the iterator itself could not printed by doing the following?

cout << it <<endl;

At first I thought the convention came from JAVA considering the security purpose. But if it is, then why I could print it's address?

However, Is there any other way to do this? If not, why?

Upvotes: 34

Views: 51372

Answers (5)

Krishnakumar T G
Krishnakumar T G

Reputation: 165

What I understand is that, you just want to know the position where the iterator points. If so you can use std::distance(numbers.begin(), it)

Upvotes: 0

Akash Ranjan
Akash Ranjan

Reputation: 1

You can't print the iterator because it is not defined to have a value. But you can perform arithematic operations on them and that helps you to print the value (of the iterator).

Upvotes: -4

user8038009
user8038009

Reputation:

Yes, there is a way to do it!

You can't print the iterator because it is not defined to have a value. But you can perform arithematic operations on them and that helps you to print the value (of the iterator).

Do the following.

cout << it - v.begin();

Example:

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

using namespace std;

int main () {
  vector<int> v = {20,3,98,34,20,11,101,201};           
  sort (v.begin(), v.end());                

  vector<int>::iterator low,up;
  low = lower_bound (v.begin(), v.end(), 20);          
  up = upper_bound (v.begin(), v.end(), 20);                  

  std::cout << "lower_bound at position " << (low - v.begin()) << std::endl;
  std::cout << "upper_bound at position " << (up - v.begin()) << std::endl;

  return 0;
}

Output of the above code:

lower_bound at position 2
upper_bound at position 4

Note: this is just a way to get things done and no way I have claimed that we can print the iterator.

...

Upvotes: 26

user5170863
user5170863

Reputation:

why the iterator itself could not printed by doing the following?

Because, it is not defined to a value internally.

Is there any other way to do this?

Basically, the compiler does not facilitate it by default, you may try to edit the compiler code! But it is too terrific you know!

If not, why?

Because it has no well-defined way to express it.

Upvotes: 0

Baum mit Augen
Baum mit Augen

Reputation: 50053

There is no predefined output operator for the standard iterators because there is no conventional meaning of printing an iterator. What would you expect such an operation to print? While you seem to expect to see the address of the object the iterator refers to, I find that not clear at all.

There is no universal answer to that, so the committee decided not to add a those operators. (The last half sentence is a guess, I am not part of the committee.)

If you want to print those iterators, I would define a function like print(Iterator); (or something like this, whatever fits your needs) that does what you want. I would not add an operator << for iterators for the reason I mentioned above.

Upvotes: 4

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