struct hack - zero sized array

Question

#include <iostream>
using namespace std;

struct node1{
    char b[3];
    int c[0];
};

struct node2{
    int c[0];
};

struct node3{
    char b[3];
};


int main() {

    cout << sizeof(node1) << endl;  // prints 4
    cout << sizeof(node2) << endl;  // prints 0
    cout << sizeof(node3) << endl;  // prints 3
}

My Question is why does the compiler allocate 0 bytes for int c[0] in node2 but allocate 1 byte for its when part of node1. I'm assuming that this 1 byte is the reason why sizeof(node1) returns 4 since without it (like in node3) its size is 3 or is that due to padding??

Also trying to understand that shouldn't node2 have enough space to hold a pointer to an array (which will be allocated in the further down in the code as part of the flexible array/struct hack?

Answer 1

Yes, it's about padding/alignment. If you add __attribute__((__packed__)) to the end [useful when writing device drivers], you'll get 3 0 3 for your output.

If node1 had defined c[1], the size is 8 not 7, because the compiler will align c to an int boundary. With packed, sizeof would be 7