Is this true for ANSI C?

Question

"The const and volatile qualifiers may precede any declaration."

I saw this statement marked as true in an online test series. But in standard C(89) I can see

declaration:
     declaration-specifiers init-declarator-listopt ;
declaration-specifiers:
     storage-class-specifier declaration-specifiersopt
     type-specifier declaration-specifiersopt
     type-qualifier declaration-specifiersopt
     function-specifier declaration-specifiersopt
init-declarator-list:
     init-declarator
     init-declarator-list , init-declarator
init-declarator:
     declarator
     declarator = initializer

which seems from above that this statement can come out false for a few declaration.

Please help!

EDIT: I know this is not valid for ISO C89 or above, but please suggest for ANSI, so that the education authority be informed about the question bug with some proof.

You did not paste the 'volatile' and 'const' from grammar. One cannot say , looking at this partial grammar. — alinsoar, Jul 23 '12 at 12:46
http://www.alc.amadeus.com/content/public/alw/skillsoft/cbtlib/116654/1101q.txt Please search the term here — perilbrain, Jul 23 '12 at 12:50
Toss that questionaire. It's full of stuff that should be *discouraged*, not *queried for*. (Reliance on specific and relative widths of generic datatypes, for example - that's what `` is for.) — DevSolar, Jul 23 '12 at 12:59

score 3 · Answer 1 · answered Jul 23 '12 at 12:50

3

type-qualifier declaration-specifiers(opt)

allows a type qualifier, such as const or volatile, followed by declaration-specifiers. Those following declaration-specifiers can be a function declaration.

answered Jul 23 '12 at 12:50

Daniel Fischer

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qualifiers may precede any declaration.... I understand that but the problem is with "any"....... – perilbrain Jul 23 '12 at 12:53
1

Well, the grammar says "any" (though, strictly, the qualifiers are part of the declaration, so they don't precede it). – Daniel Fischer Jul 23 '12 at 12:55
Of course, I thought the same, as qualifier is optional and valid with only a few type of declarations like variable declaration etc. – perilbrain Jul 23 '12 at 13:05
Of course, having a `const int foo(int n);` is pointless, the qualifier is ignored, if that's what you mean. But it's legal to have it in the declaration. – Daniel Fischer Jul 23 '12 at 13:16

bta · Answer 2 · 2013-01-05T00:17:56.843

You indeed can place const or volatile before any declaration without violating C's grammar rules. This in no way implies that such a construct has meaning, won't be ignored outright, or won't trigger a compile error for some other reason. It only means that it won't trigger a syntax error.

Section 3.5.3 of the C89 spec states

If the specification of a function type includes any type qualifiers, the behavior is undefined.

This means that it's perfectly legal to declare a function as const or volatile, as long as you don't actually call that function. If you try to call it, there's no telling what would happen. This is one of several things you can do in C that are technically legal syntax but are completely pointless (like the statement 1 == 3; or x + 2;).

To clarify your comments in your edit, be aware that the terms "ANSI C" and "C89" refer to the same thing. There are both ANSI and ISO standards for C that differ only in formatting. The content of those standards is what is commonly called "C89" or "C90" (to distinguish it from C99, which ANSI later standardized as well). When you say "I know this is not valid for ISO C89 or above", your statement includes "ANSI C" as well.

Is this true for ANSI C?

2 Answers2