Question

This is C99 code:

typedef struct expr_t
{
    int n_children; 
    foo data; // Maybe whatever type with unknown alignment
    struct expr_t *children[];
} expr_t;

Now, how do I allocate memory ?

expr_t *e = malloc (sizeof (expr_t) + n * sizeof (expr_t *));

or

expr_t *e = malloc (offsetof (expr_t, children) + n * sizeof (expr_t *));

?

Is sizeof even guaranteed to work on an type with flexible array member (GCC accepts it) ?

Answer 1

expr_t *e = malloc (sizeof (expr_t) + n * sizeof (expr_t *)); is well defined in C99. From the C99 specification 6.7.2.1.16:

As a special case, the last element of a structure with more than one
named member may have an incomplete array type; this is called a
flexible array member. In most situations, the flexible array member
is ignored. In particular, the size of the structure is as if the
flexible array member were omitted except that it may have more
trailing padding than the omission would imply.