compile error about void*

re: compile error about void*

George2 wrote:It tells you whats wrong.
Try this instead:

int main()
{
void** p;
p = new void* [100];
return 0;
}

re: compile error about void*

George2 a écrit :typedef void* PVOID;

int
main()
{
PVOID *p;
p = new PVOID[100];
return 0;
}

re: compile error about void*

George2:

Where T is a type:

If you want an array of pointers then:

T*[num]

If you want a pointer to an array, then:

T(*)[num]


--
Tomás Ó hÉilidhe

re: compile error about void*

On Jan 17, 9:28 am, Michael DOUBEZ <michael.dou...@free.frwrote:Or "p = new (void * [100]) ;".

The type specifier in a new expression has a very restricted
syntax. In particular, it cannot contain parentheses unless it
is entirely surrounded by parentheses.

Note that the syntax "new (int*)[100]" would be legal syntax,
but doesn't do what you might expect, and will invoke undefined
behavior at run-time---it allocates a single int*, then uses it
as if it were a pointer to the first element of an array,
accessing the 101st element. Of course, the resulting type of
the expression is int, so you will almost certainly get a type
error when you try to assign the results.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

re: compile error about void*

On Jan 17, 2:58 pm, "Tomás Ó hÉilidhe" <t...@lavabit.comwrote:Not in a new expression. In a new expression, you'd have to put
that in parentheses (i.e. "(T(*)[num])").

More generally, I don't think there's any context in the
language where "(void*)[100]" could be legal. If the [...] is
the subscript operator, then what precedes must be an expression
(and "(void*)" isn't a legal expression). And if the [...] is
meant to be part of a declaration, the only context in a
declaration where (void*) would be legal is as a list of
function parameters, and you can't declare a function to return
an array. (You can declare a function to return a pointer or a
reference to an array, but in this case, it would be something
like:
int (&f(void*))[100] ;
with an extra closing ) in the string.)

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

re: compile error about void*

In article <852ad1fa-f151-4bb5-b2c6-802042120aa8
@m34g2000hsf.googlegroups.com>, george4academic@yahoo.com says...You've gotten a number of answers that show the problem with the syntax
you're using. None, however, has mentioned that there are usually better
ways of doing this in C++. If you really want to do this, something like
"std::vector<void *p(100)" will do the job -- but an array (or vector)
of pointers to void sounds somewhat questionable in itself. If you're
doing something like writing your own memory allocator, this is likely
to make sense -- but for most code, a pointer to void (not to mention a
lot of pointers to void) tends to indicate a possible problem.

--
Later,
Jerry.

The universe is a figment of its own imagination.

re: compile error about void*

George2 wrote:
Wrong syntax:


The correct way to do what you want is:

void **p= new void *[100];