c - 通用字符缓冲区用作具有灵活数组成员的结构数组

Question

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

这是 this question 的 TL;DR 。仔细想想，这很有道理。

但是，可以模拟具有灵活数组成员的结构数组 - 让我们称之为swfam - 固定大小，如下所示:

#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;
}

该技巧在 C99/C11 中有效吗？我是否潜伏在未定义的行为中？

Answer 1

实现此目的的一种方法是使用指针成员而不是灵活数组。然后，您必须通过 malloc() 等手动分配其大小。等人。 [] 通常仅在数组在声明时初始化时使用，这对于 struct 是不可能的，它本质上是一个定义，而不是声明。立即声明结构类型实例的能力不会改变定义的性质，这只是为了方便。

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