How to convert from a const iterator to a non-const iterator? [Resolved]
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Thread: How to convert from a const iterator to a non-const iterator? [Resolved]

  1. #1
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    Resolved How to convert from a const iterator to a non-const iterator? [Resolved]

    Ok, here's the below snippet that's causing the problem.
    Code:
    bool graph::find_node(const std::string & node_name, const std::list<node * > & list)
    {
     for(std::list<graph::node * >::iterator iter = list.begin(); iter != list.end(); iter++)
     {
      if(( * ( * iter)).name == node_name)
      {
       return true;
      }
     }
    
     return false;
    }
    Now the list that I'm using is a constant and I would like it to remain like that in order to prevent it from being changed in any way. However, it seems that the iterator that is created from it (in the boldened line) is also a const. How could I create a non-const iterator in this instance? Or convert the const iterator to a non-const one after creation?

    [edit]

    I could change the passed in list to a non-const and it'll work, but I would not like to break good programming practices in the process, which is the reason for the question.
    Last edited by YourSurrogateGod; January 12th, 2006 at 03:48 PM.
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  2. #2
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    Post Re: How to convert from a const iterator to a non-const iterator?

    You don't need a non-const iterator. Just change the emphasized line to
    Code:
    for(std::list<graph::node * >::const_iterator iter = list.begin(); iter != list.end(); iter++)
    Last edited by googler; January 12th, 2006 at 03:54 PM.

  3. #3
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    Re: How to convert from a const iterator to a non-const iterator?

    a const_iterator means that you cannot change the object that iterator points to and this is what you want, since the list is const and hence not changeable. If you were to have a non-const iterator then you would break the contract of passing a const list.
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  4. #4
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    Re: How to convert from a const iterator to a non-const iterator?

    Why are you writing your own loop? What is wrong with std::find_if ?

    Of course, it requires a functor. You can write one thus:

    Code:
    class node_name_is
    {
       std::string searchName; // can also be const std::string & 
    public:
       node_name_is( const std::string & sn ) : searchName( sn )
       {
       }
      
        bool operator() ( graph::node * pNode )
        {
            return searchName == pNode->name;
        }
    };
    
    
    list<graph::node*> const_iterator it =
       std::find_if( nodelist.begin(), nodelist.end(),
               node_name_is( node_name ) );
    
    // check if it= nodelist.end()
    (I don't suggest you call your list list, although as long as you don't force namespace std in, there is no name-clash).

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    Re: How to convert from a const iterator to a non-const iterator?

    Quote Originally Posted by NMTop40
    Why are you writing your own loop? What is wrong with std::find_if ?

    Of course, it requires a functor. You can write one thus:

    Code:
    class node_name_is
    {
       std::string searchName; // can also be const std::string & 
    public:
       node_name_is( const std::string & sn ) : searchName( sn )
       {
       }
      
        bool operator() ( graph::node * pNode )
        {
            return searchName == pNode->name;
        }
    };
    
    
    list<graph::node*> const_iterator it =
       std::find_if( nodelist.begin(), nodelist.end(),
               node_name_is( node_name ) );
    
    // check if it= nodelist.end()
    Thanks. I couldn't find that method here, so I didn't know that it existed.
    Quote Originally Posted by NMTop40
    (I don't suggest you call your list list, although as long as you don't force namespace std in, there is no name-clash).
    That's not one of my better examples of coding skills .
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    Re: How to convert from a const iterator to a non-const iterator?

    Quote Originally Posted by NMTop40
    Why are you writing your own loop? What is wrong with std::find_if ?

    Of course, it requires a functor. You can write one thus:

    Code:
    class node_name_is
    {
       std::string searchName; // can also be const std::string & 
    public:
       node_name_is( const std::string & sn ) : searchName( sn )
       {
       }
      
        bool operator() ( graph::node * pNode )
        {
            return searchName == pNode->name;
        }
    };
    
    
    list<graph::node*> const_iterator it =
       std::find_if( nodelist.begin(), nodelist.end(),
               node_name_is( node_name ) );
    
    // check if it= nodelist.end()
    (I don't suggest you call your list list, although as long as you don't force namespace std in, there is no name-clash).
    I have another question. Why did you overload the () operator?
    Last edited by YourSurrogateGod; January 12th, 2006 at 10:04 PM.
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  7. #7
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    Re: How to convert from a const iterator to a non-const iterator?

    You should read about functors.
    here is a link http://www.newty.de/fpt/functor.html
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    Re: How to convert from a const iterator to a non-const iterator?

    Quote Originally Posted by YourSurrogateGod
    I have another question. Why did you overload the () operator?
    ...because that is how a functor or a function object is defined. The name is very intuitive - function object. It's an object that behaves syntactically like a function and to make it look so, we have to overload the "()" operator. They are superior, object oriented alternatives to function pointers.

    std::find_if is not a member of std::list. It is available under the header <algorithm>. Hope this helps. Regards.

  9. #9
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    Re: How to convert from a const iterator to a non-const iterator?

    Quote Originally Posted by exterminator
    ...because that is how a functor or a function object is defined. The name is very intuitive - function object. It's an object that behaves syntactically like a function and to make it look so, we have to overload the "()" operator. They are superior, object oriented alternatives to function pointers.

    std::find_if is not a member of std::list. It is available under the header <algorithm>. Hope this helps. Regards.
    That's cool, but why not the '==' operator? I figured that this would be more useful since the program will use comparison.

    Basically I don't understand what find_if does on the inside with the () operator.
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  10. #10
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    Re: How to convert from a const iterator to a non-const iterator?

    find_if works with what's called a predicate. Basically, a predicate takes two inputs and returns a bool. What comparisons the predicate does on the two inputs is up to it, as long as it returns a deterministic result (i.e. it always returns the same answer for the same input).

    The == operator is a predicate:
    Code:
    bool operator==(lhs, rhs)
    But, then, so is operator>, operator<, operator!= and many others. Or you can write your own (for example to compare two structs by a subset of their members).

    std::find is just std::find_if with operator== as the predicate. Or, to put another way, find_if is an extension of find that allows you to use more complicated rules for determining whether you've found what you want.
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