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

[YourSurrogateGod]

Ok, here's the below snippet that's causing the problem.

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.

[googler]

You don't need a non-const iterator. Just change the emphasized line to

for(std::list<graph::node * >::const_iterator iter = list.begin(); iter != list.end(); iter++)

[Yves M]

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.

[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:

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).

[YourSurrogateGod]

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:

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.

(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 .

[YourSurrogateGod]

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:

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?

[souldog]

You should read about functors.
here is a link http://www.newty.de/fpt/functor.html

[exterminator]

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.

[YourSurrogateGod]

...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.

[Graham]

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:

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.