7

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?

Jack
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5 Answers5

7

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.

Community
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utopianheaven
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4

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.

Ciro Santilli OurBigBook.com
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    Thanks. You're right about previous answers not quoting the right sections from the standard. The key concept was to prove that an enum value and the size of an array at the declaration time are exactly the same kind of value. – cesss Apr 21 '17 at 17:02
3

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.

2

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];
Mike
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1

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.

prazuber
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