The ANSI C grammar from -link- give me the following rules for array declarations:
(1) | direct_declarator '[' type_qualifier_list assignment_expression ']'
(2) | direct_declarator '[' type_qualifier_list ']'
(3) | direct_declarator '[' assignment_expression ']'
(4) | direct_declarator '[' STATIC type_qualifier_list assignment_expression ']'
(5) | direct_declarator '[' type_qualifier_list STATIC assignment_expression ']'
(6) | direct_declarator '[' type_qualifier_list '*' ']'
(7) | direct_declarator '[' '*' ']'
(8) | direct_declarator '[' ']'
Now I have a some questions about these:
What's the difference between the following two function prototypes:
void foo(int [*]);
and
void foo(int []);
Thank you.
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 *
.
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!)
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