understanding function object

[Wenjie]

Hello,


In the SGI STL documentation:
struct less_mag : public binary_function <double, double, bool> {
bool operator()(doub le x, double y) { return fabs(x) < fabs(y); }
};

vector<double> V;
...
sort(V.begin(), V.end(), less_mag());

I am a little confused concerning 'less_mag()': does it construct
some less_mag object? Then what should be the constructor and could
it be written as 'less_mag'? If it is calling operator(), I don't
think so :-)


Thanks for your enlightment,
Wenjie

[Alf P. Steinbach]

Re: understanding function object

On 25 Jul 2003 17:33:17 -0700, gokkog@yahoo.co m (Wenjie) wrote:
[color=blue]
>In the SGI STL documentation:
> struct less_mag : public binary_function <double, double, bool> {
> bool operator()(doub le x, double y) { return fabs(x) < fabs(y); }
> };
>
> vector<double> V;
> ...
> sort(V.begin(), V.end(), less_mag());
>
>I am a little confused concerning 'less_mag()': does it construct
>some less_mag object?[/color]

Yes.

[color=blue]
> Then what should be the constructor[/color]

It's the default constructor, which is generated by the compiler.

[color=blue]
>and could it be written as 'less_mag'?[/color]

No.

[color=blue]
>If it is calling operator(), I don't think so :-)[/color]

The sort code calls the less_mag object's operator() which takes
two arguments, the values to be compared.

[Rolf Magnus]

Re: understanding function object

Wenjie wrote:
[color=blue]
> Hello,
>
>
> In the SGI STL documentation:
> struct less_mag : public binary_function <double, double, bool> {
> bool operator()(doub le x, double y) { return fabs(x) < fabs(y); }
> };
>
> vector<double> V;
> ...
> sort(V.begin(), V.end(), less_mag());
>
> I am a little confused concerning 'less_mag()': does it construct
> some less_mag object?[/color]

Yes.
[color=blue]
> Then what should be the constructor[/color]

Since the object is empty, the constructor shouldn't do anything.
[color=blue]
> and could it be written as 'less_mag'?[/color]

No. less_mag is a type, and you cannot pass types as function arguments.
[color=blue]
> If it is calling operator(), I don't think so :-)[/color]

Well, you create an object of class less_mag, and sort() internally
calls the operator() on that object.

[Victor Bazarov]

Re: understanding function object

"Wenjie" <gokkog@yahoo.c om> wrote...[color=blue]
> In the SGI STL documentation:
> struct less_mag : public binary_function <double, double, bool> {
> bool operator()(doub le x, double y) { return fabs(x) < fabs(y); }
> };
>
> vector<double> V;
> ...
> sort(V.begin(), V.end(), less_mag());
>
> I am a little confused concerning 'less_mag()': does it construct
> some less_mag object?[/color]

Yes, a temporary object.
[color=blue]
> Then what should be the constructor and could
> it be written as 'less_mag'?[/color]

I am not sure I understand the second part of the question. The
constructor's "name" is always the same as the class'. If you need
to define a parametrised constructor, then you write

struct less_mag : ... {

less_mag(int parameter) : blahblah(blah) {}

bool operator()( ...
};
[color=blue]
> If it is calling operator(), I don't
> think so :-)[/color]

No, it's not.

Victor