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) ?
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.
If the compiler accepts the declaration of a struct with a flexible array member, the sizeof operator for that struct should yield the size of the struct as if the flexible array member doesn't exist.
The correct allocation of such a struct would be:
expr_t *e = malloc (sizeof(expr_t) + n * sizeof(struct expr_t *)); You can still do this trick even if flexible array members are not supported by the compiler. Just declare the array member of your struct as having a size of 1, and then allocate n - 1 items instead of n.
char and a float[] might be 4 bytes and require four-byte alignment, even though without the double[] it would only be one byte.
sizeof(expr_t)equals the value thatoffsetof(expr_t, children)if you had redefinedchildrento be an array of any determinate size.