在struct定义中定义宏

Question

On the definition of the structure below, there is a line with a macro definition ( #define ). What does that line do exactly? I understand it makes an alias to the first entry of the array h_addr_list , but it still looks weird to me. Is h_addr a member of the struct hostent ? Is this definition only within the scope of the struct?

struct  hostent
{
  char    *h_name;        /* official name of host */
  char    **h_aliases;    /* alias list */
  int     h_addrtype;     /* host address type */
  int     h_length;       /* length of address */
  char    **h_addr_list;  /* list of addresses from name server */
  #define h_addr  h_addr_list[0]  /* address, for backward compatiblity */
};

Answer 1

The macro definition is not scoped at all, it would work the same way if it was defined outside that struct.

It allows old code to continue using hr.h_addr rather than having to be changed to he.h_addr_list[0] (assuming he is a struct hostent ).

To clarify, h_addr is not a field of struct hostent . That define is processed by the pre-processor , the actual compiler sees an empty line where the #define is.
But if a piece of code is written as he.h_addr , the pre-processor will modify it and replace it with he.h_addr_list[0] . That is what the actual compiler will see.

Pre-processor macros are never scoped: the compiler proper doesn't even see them - it only sees the result of the substitutions, and the pre-processor ignores (is not aware of) scoping entirely.

Answer 2

Once defined, a macro identifier remains visible independently of the C scoping rules. Its scope begins at its definition, until the end of the translation unit or a corresponding #undef directive.

Here, the definition inside the structure is just for readibility; you can also define your macro after your structure for example:

struct hostent
{
  char *h_name;
  char **h_aliases;
  int h_addrtype;
  int h_length;
  char **h_addr_list;
};

#define h_addr  h_addr_list[0]

It isn't a field at all; it allows the user to write s.h_addr instead of s.h_addr_list[0] .

Answer 3

宏定义没有作用域，因此这个宏在看到此定义的编译单元中可见，直到undef h_addr 。