Question

I tried to compile something like:

struct A
{   int a;
    struct B
    {    int c;
    };
};

Now when I compile this code the compiler gives me a warning message that:

declaration does not declare anything [enabled by default]

I know that I have not defined any instance of struct B. That will mean that I shall not be able to access variable c. Still compiler compiles this code with a warning. What’s the whole point ? Why does not the compiler give a compilation error instead ?

ADDED Info:
The size of the struct A is equal to the size of int on my machine!!

Answer 1

Because you can do this:

struct A
{   int a;
    struct B
    {    int c;
    };
};

int main()
{
    struct A a = {1};
    struct B b = {2};

    return a.a + b.c;
}

Note:

• you need a semicolon after declaring B, which your code is missing
• this isn’t particularly useful, but I suppose it might serve some documentary purpose (ie,to suggest a relationship or grouping between types)
• in C++, the second variable would have type A::B, but C doesn’t have the same scoping rules (all structs just belong to the global struct namespace, in effect)

---

As to the motivation for allowing it …

struct Outer {
    struct {
        int b;
    } anon;
    /* this ^ anonymous struct can only be declared inside Outer,
       because there's no type name to declare anon with */

    struct Inner {
        int c;
    } named;
    /* this ^ struct could be declared outside, but why force it to be
       different than the anonymous one? */

    struct Related {
        double d;
    };
    /* oh no we have no member declared immediately ... should we force this
       declaration to be outside Outer now? */

    struct Inner * (*function_pointer)(struct Related *);
    /* no member but we are using it, now can it come back inside? */

    struct Related excuse;
    /* how about now? */
};

Once you’ve allowed nested type declarations like this, I doubt there’s any particular motivation to require there be a member of that type right away.