Question

Given this simple C code:

struct { 
    struct a {
        int foo;
    };
    struct b {
        char *bar;
    };
} s;

I am wondering whether there is a way to access a variable in one of the nested structures in a more compact way than s.a.foo = 5, for instance.

Answer 1

First, notice that your example is not standard C89 (but it is acceptable by some compilers when you ask for some language extensions. With GCC you’ll need to extend the accepted C dialect with the -fms-extensions flag to the compiler). You are using unnamed fields. A more standard way of coding would be:

struct a {
   int foo;
};
struct b {
   char* bar;
};
struct {
    struct a aa;
    struct b bb;
} s;

Back to your question, no, there is no other way. However, you might use preprocessor macros, whcih could help. For instance, assuming the above declarations, you could

#define afoo aa.foo
#define bbar bb.bar

and then you can code s.afoo instead of s.aa.foo

You might also define macros like

#define AFOO(X) (X).aa.foo

and then code AFOO(s)

Using such preprocessor macros does have some annoyance: with my example, you cannot declare anymore a variable (or formal argument, or field, or function) named afoo

But I am not sure you should bother. My personal advice & habit is to give longer and often unique names to fields (and also to name struct a_st my struct-ures). Take advantage of the auto-completion abilities of your editor. Don’t forget that your code is more often read than written, so use meaningful names in it.