General-purpose char buffer used as array of structs with flexible array member

Question

You can't have arrays of structures with flexible array members.

This is the TL;DR of this question. And thinking about it, it makes perfect sense.

However, may one simulate an array of structures with flexible array member - let's call them swfam - of fixed size as below :

#include <assert.h>
#include <stdlib.h>


typedef struct {
    int foo;
    float bar[];
} swfam_t; // struct with FAM

typedef struct { // this one also has a FAM but we could have used a char array instead
    size_t size,  // element count in substruct
           count; // element count in this struct
    char data[];
} swfam_array_t;


#define sizeof_swfam(size) (sizeof(swfam_t) + (size_t)(size) * sizeof(float))


swfam_array_t *swfam_array_alloc(size_t size, size_t count) {
    swfam_array_t *a = malloc(sizeof(swfam_array_t) + count * sizeof_swfam(size));

    if (a) {
        a->size = size;
        a->count = count;
    }

    return a;
}

size_t swfam_array_index(swfam_array_t *a, size_t index) {
    assert(index < a->count && "index out of bounds");
    return index * sizeof_swfam(a->size);
}

swfam_t *swfam_array_at(swfam_array_t *a, size_t index) {
    return (swfam_t *)&a->data[swfam_array_index(a, index)];
}


int main(int argc, char *argv[]) {
    swfam_array_t *a = swfam_array_alloc(100, 1000);
    assert(a && "allocation failed");

    swfam_t *s = swfam_array_at(a, 42);

    s->foo = -18; // do random stuff..
    for (int i = 0; i < 1000; ++i)
        s->bar[i] = (i * 3.141592f) / s->foo;

    free(a);
    return 0;
}

Is the trick valid C99 / C11 ? Am I lurking towards undefined behaviour ?

Anthony Bachler · Accepted Answer · 2016-10-07 19:19:47Z

0

One way to do this would be to use a pointer member instead of a flexible array. You would then have to manually allocate its size via malloc() et. al. [] is typically used only when the array is initialized on declaration, which is not possible with struct, which is essentially a definition, not a declaration. The ability to immediately declare an instance of the struct type does not change the nature of the definition, it's just for convenience.

typedef struct {
    int foo;
    float* bar; } swfam_t; // struct with FAM

answered Oct 7, 2016 at 19:19

Anthony Bachler

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