Thomas Davie wrote:
> >> I'm not familiar with your C++ example (not being familiar with C++),
> >> but I think that it's a bit of a stretch of the imagination to say
> >> that C "introduces a variable of type "array of 50 ints"", the fact
> >> that this is now an array of 50 integers is never checked at any
> >> point in the compilation or run, and I'm not sure it can be even if
> >> K&R had wanted to.
> >>
> >
> > The size is taken into account when such array type is an element of
> > another array, and by sizeof.
> >
> > int (*p)[50]; /* p may legally point only to arrays of 50 ints each */
> > ++p; /* p is assumed to point into an array, and is moved by one
> > element, i.e. by 50 ints */
> I'm not sure what you're trying to prove by saying that... There is
> still no type information that says that the contents of p are an
> array of 50 elements...
Put it this way, then:
1 void foo(void)
2 {
3 int a[2][50];
4 int b[2][60];
5 int (*p)[50];
6
7 p = a;
8 p = b;
}
$ gcc -c -Wall foo.c
foo.c: In function `foo':
foo.c:8: warning: assignment from incompatible pointer type
In line 7, an expression of type "int (*)[50]" is assigned to a
variable of type "int (*)[50]", which is OK. In line 8, an expression
of type "int (*)[60]" is assigned to a variable of type "int (*)[50]",
and the compiler complains.
> I can still attempt to access element 51 and get a runtime memory
> error.
That's because C doesn't do bounds checking on array accesses. It has
nothing to do with types.
> The type of p is still int**,
No it isn't. "int**" and "int (*)[50]" are different types and have
different run-time behaviour.
> not "pointer to array of 50 ints"
Yes it is. The semantics of C pointer arithmetic mean that the size of
the target is an essential part of the pointer type.
In C, arrays and pointers are *not* the same thing. They are often
confused because C does several automatic conversions:
1. When used as an expression (rather than an lvalue), arrays are
automatically converted to pointers. Arrays only ever occur as
lvalues, never as expressions.
2. In a declaration, the x[...] syntax indicates that x is an array,
but in an expression, x must be a pointer (which includes an array
which has been converted to a pointer due to rule 1 above).
3. When declaring function arguments, you can use "T x[]" or "T x[N]"
as an alternative syntax for "T *x"; x is still a pointer, regardless
of the syntax used.
--
Glynn Clements <[EMAIL PROTECTED]>
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