Re: Arrays

On 2024-09-29, Stefan Ram <ram@zedat.fu-berlin.de> wrote:

  Alright, so I've got a triple whammy of questions coming at me.
  It's about this program I've been tinkering with.
>
#include <stdio.h>
int main()
{ char const abc[] = "abcdefhijklmnopqrstuvwxyz";
  char const * const p = abc + 3;
  char const( * const def )[] =( char const( * const )[] )p;

                                  ^^^^^^^^^^^^^^^^^^^^^^^^^
                                  ]
This is a const pointer to an array (of unknown size) to const char.

  printf( "%c\n",( *def )[ -3 ]); }
>
  1st
>
  First off, I'm getting this warning that's throwing me for a loop:
>
main.c: In function 'main':
main.c:7:32: warning: cast discards 'const' qualifier from pointer target type [-Wcast-qual]
    7 |   char const( * const def )[] =( char const( * const )[] )p;

The problem is that def is a pointer to an array of const char.

The const char elements are const qualified, but the array type is not!

That's why a qualifier is being stripped in the cast.

  So I'm wondering: How can I dodge this discarding the constness
  warning?

I would say, don't work with pointers to arrays.

There is no way to declare a const array in declarator syntax.
You must create a typedef name N for the array type and then
refer to it as const N.

  2nd
>
  Then I'm hit with another warning:
>
main.c:7:3: warning: dereferencing type-punned pointer might break strict-aliasing rules [-Wstrict-aliasing]
    7 |   char const( * const def )[] =( char const( * const )[] )p;
      |   ^~~~
>
  What's this warning trying to tell me?

That you're banging your head against array-pointer-const corner cases
in the C language.

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