Anonymous union within struct not in c99?

Question

here is very simplified code of problem I have:

enum node_type {
    t_int, t_double
};

struct int_node {
    int value;
};

struct double_node {
    double value;
};

struct node {
    enum node_type type;
    union {
        struct int_node int_n;
        struct double_node double_n;
    };
};

int main(void) {
    struct int_node i;
    i.value = 10;
    struct node n;
    n.type = t_int;
    n.int_n = i;
    return 0;
}

And what I don't undestand is this:

$ cc us.c 
$ cc -std=c99 us.c 
us.c:18:4: warning: declaration does not declare anything
us.c: In function ‘main’:
us.c:26:4: error: ‘struct node’ has no member named ‘int_n’

Using GCC without -std option compiles code above without any problems (and the similar code is working pretty well), but it seems that c99 does not permit this technique. Why is it so and is it possible to make is c99 (or c89, c90) compatible? Thanks.

Solution 1

Anonymous unions are a GNU extension, not part of any standard version of the C language. You can use -std=gnu99 or something like that for c99+GNU extensions, but it's best to write proper C and not rely on extensions which provide nothing but syntactic sugar...

Edit: Anonymous unions were added in C11, so they are now a standard part of the language. Presumably GCC's -std=c11 lets you use them.

Solution 2

I'm finding this question about a year and a half after everybody else did, so I can give a different answer: anonymous structs are not in the C99 standard, but they are in the C11 standard. GCC and clang already support this (the C11 standard seems to have lifted the feature from Microsoft, and GCC has provided support for some MSFT extensions for some time).

Solution 3

Well, the solution was to name instance of the union (which can remain anonymous as datatype) and then use that name as a proxy.

$ diff -u old_us.c us.c 
--- old_us.c    2010-07-12 13:49:25.000000000 +0200
+++ us.c        2010-07-12 13:49:02.000000000 +0200
@@ -15,7 +15,7 @@
   union {
     struct int_node int_n;
     struct double_node double_n;
-  };
+  } data;
 };

 int main(void) {
@@ -23,6 +23,6 @@
   i.value = 10;
   struct node n;
   n.type = t_int;
-  n.int_n = i;
+  n.data.int_n = i;
   return 0;
 }

Now it compiles as c99 without any problems.

$ cc -std=c99 us.c 
$ 

Note: I am not happy about this solution anyway.

Solution 4

Just for clarifications about anonymous struct or anonymous union.

C11

6.7.2.1 Structure and union specifiers

An unnamed member whose type specifier is a structure specifier with no tag is called an anonymous structure; an unnamed member whose type specifier is a union specifier with no tag is called an anonymous union. The members of an anonymous structure or union are considered to be members of the containing structure or union. This applies recursively if the containing structure or union is also anonymous.

C99 There are no anonymous struct or union

Simplified: Type-specifier Identifier { Declaration-list } Tags ;

• Type-specifier: struct or union;
• Identifier: optional, your custom name for the struct or union;
• Declaration-list: members, your variables, anonymous struct and anonymous union
• Tags: optional. If you have a typedef in front of the Type-specifier, the Tags are alias and not Tags.

It is a anonymous struct or anonymous union only if it have no identifier and no tag, and exist inside another struct or union.

struct s {
    struct { int x; };     // Anonymous struct, no identifier and no tag
    struct a { int x; };   // NOT Anonymous struct, has an identifier 'a'
    struct { int x; } b;   // NOT Anonymous struct, has a tag 'b'
    struct c { int x; } C; // NOT Anonymous struct
};

struct s {
    union { int x; };     // Anonymous union, no identifier and no tag
    union a { int x; };   // NOT Anonymous union, has an identifier 'a'
    union { int x; } b;   // NOT Anonymous union, has a tag 'b'
    union c { int x; } C; // NOT Anonymous union
};

typedef hell: if you have a typedef the tag part is not a tag anymore, it is alias for that type.

struct a { int x; } A; // 'A' is a tag
union a { int x; } A;  // 'A' is a tag

// But if you use this way
typedef struct b { int x; } B; // 'B' is NOT a tag. It is an alias to struct 'b'
typedef union b { int x; } B;  // 'B' is NOT a tag. It is an alias to union 'b'

// Usage
A.x = 10; // A tag you can use without having to declare a new variable

B.x = 10; // Does not work

B bb; // Because 'B' is an alias, you have to declare a new variable
bb.x = 10;

The example bellow just change struct for union, work the same way.

struct a { int x; }; // Regular complete struct type
typedef struct a aa; // Alias 'aa' for the struct 'a'

struct { int x; } b; // Tag 'b'
typedef struct b bb; // Compile, but unusable.

struct c { int x; } C; // identifier or struct name 'c' and tag 'C'
typedef struct { int x; } d; // Alias 'd'
typedef struct e { int x; } ee; // struct 'e' and alias 'ee'

Solution 5

Looking at 6.2.7.1 of C99, I'm seeing that the identifier is optional:

struct-or-union-specifier:
    struct-or-union identifier-opt { struct-declaration-list }
    struct-or-union identifier

struct-or-union:
    struct
    union

struct-declaration-list:
    struct-declaration
    struct-declaration-list struct-declaration

struct-declaration:
    specifier-qualifier-list struct-declarator-list ;

specifier-qualifier-list:
    type-specifier specifier-qualifier-list-opt
    type-qualifier specifier-qualifier-list-opt

I've been up and down searching, and cannot find any reference to anonymous unions being against the spec. The whole -opt suffix indicates that the thing, in this case identifier is optional according to 6.1.

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