>
> You're really tying yourself in knots here. The
> address of an array
> is the address of its first member, so
>
> int main() {
>
> int array[9] = {0};
>
> const int *p = array;
>
> return p[0];
> }
>
> Andrew.
>
The advantage of declaring a pointer like I had mentioned in my original
posting is that I will get a compiler error if someone tries to give me a
buffer of a different size. Let's say you are only working
with fixed size buffers and you are expecting to get a pointer to one of
them. The prototype of your function could look like this:
void foo(const int * p, int plen);
However, in this case you will have to:
a) Verify that plen matches the fixed size you are expecting
b) Trust that the caller indeed gave you a buffer p with valid plen
memory locations.
c) Problems will not be caught at compile time. Only at run-time.
This is ok if you are working with buffers that may vary in size. But if
your buffers have a fixed size, wouldn't it be better to do this?
void foo(const int (* p)[9]);
In this case, you are guaranteed that the caller is giving you a valid
buffer. Otherwise, you will get a compile error message which will help
you quickly detect a problem.
Do you understand now the motivation behind the question?
This is the reason why I am trying to do this:
int array[9];
const int (* p)[9] = &array;
But gcc complains that:
"warning: initialization from incompatible pointer type"
And I don't see why. Where in the standard it says that I am not allowed
to do this? Should I log a bug against gcc so that I can get an explanation
from the developers themselves?
Thanks
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