From:

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <Erik-wikstrom@telia.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 15 Oct 2007 19:50:47 GMT

Message-ID:

<rwPQi.11416$ZA.7619@newsb.telia.net>

This is my first post here, please tell me if I did anything wrong.

in the following code snippet:

1. template <class In, class Pred>

2. In find_if(In begin, In end, Pred f) {

3. while (begin != end && !f(*begin))

4. ++begin;

5. return begin;

6. }

7. bool is_negative(int n) {

8. return n < 0;

9. }

10. vector<int>::iterator i = find_if(v.begin(), v.end(),

is_negative);

find_if is the template needs a class Pred as its third argument but

in line 10's function call, it takes a function poniter is_negative,

does that mean in C++, function point is equivalent to class?

in the following code snippet:

1. template <class In, class Pred>

2. In find_if(In begin, In end, Pred f) {

3. while (begin != end && !f(*begin))

4. ++begin;

5. return begin;

6. }

7. bool is_negative(int n) {

8. return n < 0;

9. }

10. vector<int>::iterator i = find_if(v.begin(), v.end(),

is_negative);

find_if is the template needs a class Pred as its third argument but

in line 10's function call, it takes a function poniter is_negative,

does that mean in C++, function point is equivalent to class?

No. If you rewrite the template thingie like this instead

template <typename In, typename Pred>

In find_if(In begin, In end, Pred f)

it might become a little more clear, the types of In and Pred are

determined when the code is compiled, and it does not have to be a

class, anything that is a type will do.

If we take you code and add (before line 10) the following:

struct IsNegative {

bool operator()(int n) { return n < 0; }

};

And after line 10 we add

IsNegative pred;

vector<int>::iterator i2 = find_if(v.begin(),v.end(), pred);

Now, in the first call the type of Pred in find_if() was a pointer to a

function, while in the second call the type is IsNegative. This process

where the types of the parameters are decided is called instantiation.

The template is not code that can be executed, it is just a template

describing a generic algorithm (or class), to be useful we must use this

description and fill in the parameters with real types, which created a

real function (or class). So with the additions that I made your program

will now contain two find_if functions, the first has the following

signature:

typedef vector<int>::iterator iter;

iter find_if(iter, iter, bool (*)(int));

Notice that the last argument is a function pointer. The second find_if

function have this signature:

iter find_if(iter, iter, IsNegative);

The last argument is now an instance of the IsNegative class.

Hope this clear some things up for you.

--

Erik Wikstr??m

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