枚举成员可以是ANSI-C中数组的大小吗？

Question

I need to allocate an array according to how many elements the enum have. I did the following:

enum { A, B, C, LAST };
char buf[LAST];

That works fine,even with -ansi -pedantic flags. But I'm not sure if it's a GCC or clang(wich supports most,if not all GCC-extensions) extensions or really allowed by the ANSI C standard and will works fine in any C compiler with ANSI-C std. Can someone clarify it?

Answer 1

Both the C89 (section 3.5.2.2) and C99 (section 6.7.2.2) standards define enums the same way:

6.7.2.2 Enumeration specifiers (Paragraph 3), http://www.open-std.org/JTC1/SC22/WG14/www/docs/n1256.pdf

3.5.2.2 Enumeration specifiers (Paragraph 3), http://flash-gordon.me.uk/ansi.c.txt

Both read:

[...] An enumerator with = defines its enumeration constant as the value of the constant expression. If the first enumerator has no =,the value of its enumeration constant is 0. Each subsequent enumerator with no = defines its enumeration constant as the value of the constant expression obtained by adding 1 to the value of the previous enumeration constant. [...]

Therefore, in your syntax, any standard-compliant compiler will run your code correctly.

Answer 2

That works fine,even with -ansi -pedantic flags

So it's not a GNU extension. Yes, this is fine in ANSI C, because members of an enum are constant expressions.

Answer 3

As others have said, it is valid. But I think no one has quoted the right sections so far. The relevant ones from the N1256 C99 draft are: 6.6 "Constant expressions" paragraph 6:

An integer constant expression99) shall have integer type and shall only have operands that are integer constants, enumeration constants [...]

and then 6.7.5.2 "Array declarators" paragraph 4:

If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type [...]

So basically:

• enumeration constants are constant expressions
• for the array not to be variable length, we need a constant expression

I believe that 6.7.2.2 "Enumeration specifiers" which others quoted talks about declaring the enum , not using the enumerators. Of course, since when declaring them you need compile time constants, we expect that they should also be compile time constants when used in expressions.

Answer 4

Can someone clarify it?

I'm sure you know, an enum is just appling a label to a number:

enum
{ A,  // 0 
  B,  // 1
  C,  // 2
  LAST  // 3
};

So really:

char buf[LAST];

Is no different than:

char buf[3];

Answer 5

From C Standard, paragraph 6.2.5 (Types):

16 An enumeration comprises a set of named integer constant values. Each distinct enumeration constitutes a different enumerated type.

17 The type char, the signed and unsigned integer types, and the enumerated types are collectively called integer types.

Also, paragraph 6.7.2.2 (Enumeration specifiers):

The expression that defines the value of an enumeration constant shall be an integer constant expression that has a value representable as an int.