Question

Reading this Skip List implementation I came across this code fragment:

typedef struct nodeStructure{
    keyType key;
    valueType value;
    node forward[1]; /* variable sized array of forward pointers */
    };

To me it seems that forward[1] denotes a one element array. And the comment calls it a variable sized array.

Do I misunderstand something or this is just a mistake in the source I’m reading?

Answer 1

This is a common trick for the older C compilers (before C99): compilers allowed you to dereference elements past the end of forward‘s declared length when it is the last element of the struct; you could then malloc enough memory for the additional node elements, like this:

nodeStructure *ptr = malloc(sizeof(nodeStructure)+4*sizeof(node));
for (int i = 0 ; i != 5 ; i++) { // The fifth element is part of the struct
    ptr->forward[i] = ...
}
free(ptr);

The trick lets you embed arrays of variable size in a structure without a separate dynamic allocation. An alternative solution would be to declare node *forward, but then you’d need to malloc and free it separately from the nodeStructure, unnecessarily doubling the number of mallocs and potentially increasing memory fragmentation:

Here is how the above fragment would look without the hack:

typedef struct nodeStructure{
    keyType key;
    valueType value;
    node *forward;
};

nodeStructure *ptr = malloc(sizeof(nodeStructure));
ptr->forward = malloc(5*sizeof(node));
for (int i = 0 ; i != 5 ; i++) {
    ptr->forward[i] = ...
}
free(ptr->forward);
free(ptr);

EDIT (in response to comments by Adam Rosenfield): C99 lets you define arrays with no size, like this: node forward[]; This is called flexible array member, it is defined in the section 6.7.2.1.16 of the C99 standard.