Can someone explain why this doesn’t work? Below is a simple example of what I’m trying to do. I don’t need an explanation on error: flexible array member. I understand that if int g[] is not at the end of the struct, there is ambiguity about the size of the struct that needs to be allocated. In my mind though, creating a union with a particular size seems to get around this. The union here is 100 bytes, so if I use the union, then there shouldn’t be ambiguity about the structs size.

My expectation is that the struct S, by default will be evaluated as 204 bytes.

I’m dealing with this problem because I’m using some C code in a C++ project. When the C code is compiled in a C context, this code compiles and works as I’ve described. But in C++, I get the error: flexible array member error.

I’m trying to explore all options before refactoring the C code and I’m trying to understand why the C++ compiler doesn’t allow this while the C compiler does.

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

struct B {
    int d;
    int e;
    int f;
    int g[];
};

union U {
    B b;
    A a;
    uint8_t arr[100];
};

struct S {
    int b;
    U u1;
    U u2;
};

int main()
{
    U u = {{0}};
    S s = {0};

    printf("size=%lun", sizeof(U));

    return 0;
}

error

error: flexible array member ‘B::g’ not at end of ‘struct S’