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Answer 1

A:

Point2D needs to inherit from std::iterator before it's going to place nicely with std::distance. std::distance needs some of the typedefs provided by the standard iterator type in order to use the correct method for measuring the distance.

EDIT: I'm assuming Point2D is supposed to be an iterator. If you're assuming std::distance is the distance formula, you are mistaken. std::distance returns the difference in position between two iterators pointing to a common container.

Billy ONeal 2010-04-23 06:28:00

I did not understand anything. Please elaborate. The code looks perfectly fine to me then why this strange error?

avd 2010-04-23 06:29:30

Okay you are saying that distance is already a function in standard library?

avd 2010-04-23 06:30:52

Yes, `std::distance` is already part of the standard library: http://www.cplusplus.com/reference/std/iterator/distance/ That's why it's in namespace std, so you can define your own function called distance if you want. You don't have any `using namespace std` s around, do you? :)

Billy ONeal 2010-04-23 06:33:55

@GMan: Ow! Get off my toes! (And maybe I'll get off yours)

Billy ONeal 2010-04-23 06:35:07

@Billy: Fine. :P

GMan 2010-04-23 06:35:49

@GMan: LOL! You didn't have to *delete* it. *Bill is cracking up*

Billy ONeal 2010-04-23 06:38:03

No actually I have namespace std. I got it.

avd 2010-04-23 06:39:05

@Billy ONeal: It doesn't matter whether the standard `distance` is in `std::` or not. C++ supports function overloading, based on argument types. If the code is indeed as the OP describes it, the call in the OP must resolve to the OP's `distance`, not to `std::distance`. `using namespace std` cannot affect that in any way.

AndreyT 2010-04-23 06:48:36

@AndreyT: That's not what the OP is claiming. Honestly, when I wrote this answer, I paid more attention to the error message than the code (Which is evident in the first part of the answer :) ) I don't know.

Billy ONeal 2010-04-23 06:58:23

Answer 2

A:

Looks like you may be calling std::distance instead of your own distance function. There are a few reasons why this might be happening, but it's usually because of gratuitous use of

using namespace std;

Try explicitly using the namespace or class name when calling the distance function, or if it's in the global namespace

float result = ::distance(point1, point2);

tjohns20 2010-04-23 06:34:57

Or (best option): Get rid of `using namespace std;`. It makes code in books look prettier, but there's a reason the standard library is in a namespace.

Billy ONeal 2010-04-23 06:40:07

Answer 3

A:

Maybe compiler interprets distance call as std::distance? Do you have using namespace std before this code? Try to rename the distance function and see what happens.

Alex Farber 2010-04-23 06:35:38

I think it is obvious that the compiler interprets the call as `std::distance`. No doubt about that. Yet I don't see how it can interpret it that way given the above code.

AndreyT 2010-04-23 06:45:35

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