Re: "span"

On 2024-03-23, Ben Bacarisse <ben.usenet@bsb.me.uk> wrote:

ram@zedat.fu-berlin.de (Stefan Ram) writes:
>

  Some people suggested to introduce a feature "span" (probably a
  type) into C++, so that when one has an int array "a" and calls
  "f( a )", the length of a is being transferred to "f", which "f"
  is defined using "void f(span<int> a)". The "span" in "f" then
  knows the length of "a". This is supposed to be less error prone
  than passing a pointer with a separate length argument.
>
  Of course, I immediately wondered whether one could implement
  such a "span" for C, and here's a draft:
>
#include <stdio.h>
>
#define SPAN_PARAM(x,a,n) x*a,size_t const n
#define SPAN(a) a,(sizeof a/sizeof 0[a])

>
The usual reason given for writing 0[a] in a macro is so that
parentheses are not needed as they would be (recommended) for *(a) or
(a)[0].  But since you don't seem to care about writing a rather than
(a) I don't see why you wrote 0[a] rather than either *a or a[0].
>

void print( SPAN_PARAM( int, a, n ))
{ for( size_t i = 0; i < n; ++i )
  printf( "%lld: %d\n", i, a[ i ]); }

>
What happens when print wants to pass 'a' to another function that takes
a span?

How about linking the names of a and the size, like a and a_n?
Then we can make sure there is a _n size, whether we have an
array or pointer:

  #include <stddef.h>
  #include <stdio.h>

  #define SPAN_DEF(type, a, size, init)                 \
    type a[size] = init; const size_t a ## _n = size

  #define SPAN_PARAM(type, a)                           \
    type *a, size_t const a ## _n

  #define SPAN(a) a, a ## _n

  void print(SPAN_PARAM(int, a))
  {
    for (size_t i = 0; i < a_n; i++)
      printf("%zd, %d\n", i, a[i]);
  }

  void print_wrap(SPAN_PARAM(int, a))
  {
    print(SPAN(a));
  }

  int main(void)
  {
    SPAN_DEF(int, a, 42, { 0 });
    print_wrap(SPAN(a));
  }

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