How can you declare function f that...

[benben]

How do you declare a function f that takes a parameter that is a pointer
to itself?

So you can do f(f);

Pardon my curiosity!

Ben

 

[Vyacheslav Kononenko]

How do you declare a function f that takes a parameter that is a pointer
to itself?

So you can do f(f);

Pardon my curiosity!

Ben

struct Boo { template<class T>Boo( T ); };

void f( Boo );

void foo()
{
f( f );
}

 

[softcoder]

Pardon my curiosity! but can anybody please explain why this code
works?

-softcoder

 

[John Carson]

Pardon my curiosity! but can anybody please explain why this code
works?

-softcoder

Does it? At least on my compiler, it fails to link. This is easily corrected
with a couple of empty definitions, but it still doesn't give the results
one might want.

struct Boo { template<class T>Boo( T ); };

which I would re-write as:

struct Boo
{
template<class T>
Boo( T );
};

gives Boo a templated constructor. You can pass Boo's constructor an
argument of any type and a constructor that takes that type will be produced
by the template mechanism. Thus, when you pass it f, the template parameter
T becomes f.

void f( Boo );

defines f to accept a Boo parameter. Thus when

f(f);

is called, f is converted to a Boo temporary using Boo's templated
constructor and this Boo temporary is passed to f (or at least it would be
if the compiler didn't optimise it away).

Accordingly, while f is used as the function argument, the f function never
gets to see this argument, which rather defeats the purpose (at least you
would think so; the OP didn't explain the purpose beyond satisfying his
curiousity).

Observe that recursive functions call themselves without needing a function
pointer to themselves, so it is not clear what the point of

f(f);

might be.

With function objects, you can achieve something that "works" after a
fashion:

struct F
{
F() : count(0)
{}
void operator()(F& f)
{
cout << "f\n";
++count;
if(count < 10)
f(*this);
}
int count;
};

int main()
{
F f;
f(f);

return 0;
}

 

[Lionel B]

How do you declare a function f that takes a parameter that is a pointer
to itself?

So you can do f(f);

struct ftype
{
void operator() (ftype) const {}
};

int main() {
ftype f;
f(f);
}

Well ok, so functor, not function... and it wasn't actually supposed to
do anything was it?

 

[benben]

struct ftype
{
void operator() (ftype) const {}
};

int main() {
ftype f;
f(f);
}

Well ok, so functor, not function... and it wasn't actually supposed to
do anything was it?

No it's just I am a bit curios about this, nothing big deal at all. I'm
thinking something like

void f( PTR_TO_FUNC inner)
{
if (inner == 0)
return;

if (...) inner(0);
else inner(inner);
}

Ya I know it's trivial with functor, but I'm just so curios if anyone
can do it with a straight function

Ben