ANSI-C grammar - array declarations like [*] et alii

Question

The ANSI C grammar from -link- give me the following rules for array declarations:

 (1) | direct_declarator \'[\' type_qualifier_list assignment_expression \']\'         


        

Answer 1

You can't use type qualifiers or static in size portion of array declaration in C89/90. These features are specific to C99.

static in array declaration tells the compiler that you promise that the specified number of elements will always be present in the array passed as the actual argument. This might help compilers to generate more efficient code. If you violate your promise in the actual code (i.e. pass a smaller array), the behavior is undefined. For example,

void foo(int a[static 3]) {
  ...
}

int main() {
  int a[4], b[2];
  foo(a); /* OK */
  foo(b); /* Undefined behavior */
}

The * in size portion of array declaration is used in function prototype declarations only. It indicates that the array has variable length (VLA). For example, in the function definition you can use a VLA with a concrete run-time size

void foo(int n, int a[n]) /* `a` is VLA because `n` is not a constant */
{
  ...
}

When you declare the prototype you can do the same

void foo(int n, int a[n]); /* `a` is VLA because `n` is not a constant */

but if you don't specify the parameter names (which is OK in the prototype), you can't use n as array size of course. Yet, if you still have to tell the compiler that the array is going to be a VLA, you can use the * for that purpose

void foo(int, int a[*]); /* `a` is VLA because size is `*` */

Note, that the example with a 1D array is not a good one. Even if you omit the * and declare the above function as

void foo(int, int a[]);

then the code will still work fine, because in function parameter declarations array type is implicitly replaced with pointer type anyway. But once you start using multi-dimensional arrays, the proper use of * becomes important. For example, if the function is defined as

void bar(int n, int m[n][n]) { /* 2D VLA */
  ...
}

the the prototype might look as follows

void bar(int n, int m[n][n]); /* 2D VLA */

or as

void bar(int, int m[*][*]); /* 2d VLA */

In the latter case the first * can be omitted (because of the array-to-pointer replacement), but not the second *.

Answer 2

My K&R2nd (which covers and includes the ANSI standard) does not seem to say anything about [*] either in the text, or in the standard itself. Nor can I make the official grammar in the standard accept that syntax.

It may be related to K&R c (though I don't seem to recall it), may have been a common extension, or have been a proposal that ultimately didn't make the standard.

I would assume it makes the dimension of the array explicitly unspecified. But I'm just guessing.

---

Hmm...gcc accepts

#include <stdio.h>

void f(int s, int a[*]);

int main(void){
  int a[2] = {0};
  f(2,a);
  return 0;
}

void f(int s, int a[]){
  int i;
  for (i=0; i<s; ++i){
    printf("%d\n",a[i]);
  }
}

in ansi, c89 and c99 mode; issuing no warnings even with -Wall. Note that it did not like the [*] syntax in the function definition. Adding -pedantic made it complain about the [*] syntax in c89 and ansi modes, but it continued to accept in in c99 mode.

Answer 3

I hope you are not trying to learn C grammar from a yacc specification!? The link you posted appears to be based on the ISO C99 draft. The relevant section is 6.7.5.2. The wording is arcane (but less so than the yacc syntax perhaps!)