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Re: [PATCH] warning about const multidimensional array as function parameter


Jonathan Wakely <jwakely.gcc@gmail.com>:

> On 27 October 2014 13:10, Joseph S. Myers wrote:
> > On Sat, 25 Oct 2014, Martin Uecker wrote:
> >
> >> Strictly speaking the C standard considers such pointers to be
> >> incompatible. This seems to be an unintentional consequence
> >> of how qualifiers are always attached to the element type.
> >> (I am trying to get the standard revised too.) The new
> >> behaviour should also be more compatible with C++.
> >
> > What is the exact difference in wording in the C++ standard that results
> > in this difference in semantics?
> 
> See 4.4 [conv.qual] in https://isocpp.org/files/papers/N3797.pdf

Note that this doesn't talk explicitly about arrays
and that C++ keeps the notion that qualifiers are
always attached to the element type:

---
3.9.3(2) ... Any cv-qualifiers applied to an array type
affect the array element type, not the array type (8.3.4).
---

and

---
3.9.3(5) ...
Cv-qualifiers applied to an array type attach to the
underlying element type, so the notation âcv T,â where
T is an array type, refers to an array whose elements
are so-qualified. Such array types can be said to be
more (or less) cv-qualified than other types based on
the cv-qualification of the underlying element types.
---

I *believe* (but I don't know the C++ standard very well)
that all the magic is in the wording "can be said to be more 
(or less) cv-qualified" which makes the conversion rules work 
for arrays with constant element type "as if" the array itself
had the qualifier.

---
4.4 Qualification conversions
4.4(1) "A prvalue of type âpointer to cv1 Tâ can be converted
to  a prvalue of type âpointer to cv2 Tâ if âcv2 Tâ is more
cv-qualified than âcv1 Tâ."
---


There is another issue in C which has the same underlying reason
(brought up by Tim Rentsch in comp.std.c) as shown in the following
example (this is legal C and compiles without a warning (gcc) but is
illegal in C++):

#include <string.h>
static const int x[5] = { 0 };
void test(void)
{
        memset(&x, 0, sizeof(x));
}

I did not try to address this in the patch because it would make legal
code have a warning, but one could think about it.


Martin

PS: Joseph, thank you for reviewing the patch.


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