can you return a void function call?

[deanbrown3d]

Hi there!

Suppose I have a function

void ShowMessage(str ing s);

(that has a void return type.)
In my other function F, also with a void return type, can I do this?

void F(int i)
{
// Can I return this?
return ShowMessage("Hi Dean"); // ????

}

I can seem to be able to do this in Borland Builder6, but I didn't
think it was possible until recently. Is this a compiler-specific
setting?
THX!

Dean

[Gary Labowitz]

<de*********@ya hoo.com> wrote in message
news:11******** **************@ l41g2000cwc.goo glegroups.com.. .

Hi there!

Suppose I have a function

void ShowMessage(str ing s);

(that has a void return type.)
In my other function F, also with a void return type, can I do this?

void F(int i)
{
// Can I return this?
return ShowMessage("Hi Dean"); // ????

}

I can seem to be able to do this in Borland Builder6, but I didn't
think it was possible until recently. Is this a compiler-specific
setting?

No, this doesn't make sense. It is invalid to return a value from a function
returning void.
--
Gary

[deanbrown3d]

Even if the value is void?

[deanbrown3d]

I mean, it seems to make sense to say:

return ShowMessage("Hi "); // One line

then to write:

{
ShowMessage("Hi "); // Four lines
return;
}

[Rolf Magnus]

Gary Labowitz wrote:

In my other function F, also with a void return type, can I do this?

void F(int i)
{
// Can I return this?
return ShowMessage("Hi Dean"); // ????

}

I can seem to be able to do this in Borland Builder6, but I didn't
think it was possible until recently. Is this a compiler-specific
setting?

No, this doesn't make sense. It is invalid to return a value from a
function returning void.

Nonsense! It is perfectly valid to do that, as long as the 'value' is of
type void:

void a()
{
}

void b()
{
return a();
}

The reason for this being allowed are templates. Think e.g. of the standard
library's mem_fun, which might use something like this (copied from
TC++PL):

template <class R, class T> class mem_fun_t : public unary_function< T*, R> {
R(T::*pmf)();
public:
explicit mem_fun_t(R(T:: *p)()) : pmf(p) {}
R operator()(T* p) const { return (p->*pmf)(); }
};

Now if you have a member function returning void:

class X
{
public:
void foo() {};
};

int main()
{
std::mem_fun_t< void, X> fun = &X::foo;
fun(&X);
}

for 'fun', the operator becomes:

void operator()(X* p) const { return (p->*pmf)(); }

And (p->*pmf) becomes a function returning void.

[Matthias Kaeppler]

de*********@yah oo.com wrote:

Even if the value is void?

void is not a value, void is nothing, so it doesn't make sense to return
anything. If you want to return something, then you'll have to use the
right signature for your function.
Invoking 'return' without passing a value just means that you are about
to leave the scope of the function.

--
Matthias Kaeppler

[deanbrown3d]

Wow and there's the reason! Thanks Rolf, makes sense. It must have
always been that way, I just didn't notice.

Dean

[Gary Labowitz]

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:d0******** *****@news.t-online.com...

Gary Labowitz wrote:

In my other function F, also with a void return type, can I do this?

void F(int i)
{
// Can I return this?
return ShowMessage("Hi Dean"); // ????

}

I can seem to be able to do this in Borland Builder6, but I didn't
think it was possible until recently. Is this a compiler-specific
setting?

No, this doesn't make sense. It is invalid to return a value from a
function returning void.

Nonsense! It is perfectly valid to do that, as long as the 'value' is of
type void:

void a()
{
}

void b()
{
return a();
}

Interesting.
If you coded b()'s return as
return void; //compiler error
But with the call to a() returning nothing, it looks like the 'nothing' is
the temporary value as a result of evaluating a(). On the other hand, the
compiler may be optimizing out the call to a( ) since a is empty. But no, if
there is code in the a( ) body it executes and so does the return.
void is surely tricky business, and the reason to allow void as a 'return'
from a function call makes sense.
There would appear to be no other way to set a temporary to void, I should
think, but if there is I'd like to see it.
--
Gary

[Howard]

"Gary Labowitz" <gl*******@comc ast.net> wrote in message
news:zp******** ************@co mcast.com...

Interesting.
If you coded b()'s return as
return void; //compiler error
That's an error because "void" is a type, and you don't return types, you
return objects. It's like writing:
return int; // compiler error
There would appear to be no other way to set a temporary to void, I should
think, but if there is I'd like to see it.

I can't think of any way, either. I tried using something like
DoVoid(DoVoid() ), and even DoVoid((void)(D oVoid())), but that doesn't
compile. (CodeWarrior just says the signatures don't match, but doesn't say
what it thinks the parameter is, if not a void temporary.)

-Howard

[Rolf Magnus]

Howard wrote:

"Gary Labowitz" <gl*******@comc ast.net> wrote in message
news:zp******** ************@co mcast.com...

Interesting.
If you coded b()'s return as
return void; //compiler error

That's an error because "void" is a type, and you don't return types, you
return objects. It's like writing:
return int; // compiler error

You can however do:

return (void)0;

Or (at least on g++):

return void();

There would appear to be no other way to set a temporary to void, I
should think, but if there is I'd like to see it.

I can't think of any way, either. I tried using something like
DoVoid(DoVoid() ), and even DoVoid((void)(D oVoid())), but that doesn't
compile. (CodeWarrior just says the signatures don't match, but doesn't
say what it thinks the parameter is, if not a void temporary.)

g++ says "too many arguments", since the function expects zero argument, but
I provided one. Even though it's of type void, it is nevertheless an
argument. It would be a nice thing to have void parameters. This could be
useful in templates.