the general usage of flexible array is to use malloc to define the flexible array. I'm trying to explore defining the flexible array with another struct. An example
typedef struct {
uint64_t header;
size_t data_size;
float data[];
} test_base_t;
typedef struct {
test_base_t base;
float data[3];
} test_t;
As I understand, flexible array needs to be defined at the end of a struct. And clangd will give the following warning. -Wgnu-variable-sized-type-not-at-end
I just wanted to ask if anybody has done this before and is it safe? Or is there a better way to define flexible array size without alloc?
You can then wrap the usage of the object in a macro to static assert ext.base.data == ext.data
before casting and passing to a general API consumes test_base_t
. This way you can have the memory required in compile instead of allocing.
Edit
There seem to be a confusion on how I wanted to consume it, here is an example to demonstrate
#define SUM_BASE(test) \
sum_base(&test->base); \
_Static_assert(test->data == test->base.data);
float sum_base(test_base_t *base)
{
float sum = 0;
for (size_t i = 0; i < base->data_size; i )
{
sum = base->data[i];
}
return sum;
}
test_t test = { .base = { .data_size = 3, }, .data = { 1, 2, 3, }, };
SUM_BASE((&test));
CodePudding user response:
You cannot create actual instances of test_base_t
with an initialized array, but you can create compound literals with an initialized array of a specified length and cast their address as test_base_t
pointers. The layout and alignment of both structures should be compatible, given that they have exactly the same types, save for the flexible array length.
Here is an example:
#include <inttypes.h>
#include <stdio.h>
#include <stdint.h>
typedef struct {
uint64_t header;
size_t data_size;
float data[];
} test_base_t;
#define TEST_ARRAY(n) (test_base_t*)&(struct { uint64_t header; size_t data_size; float data[n]; })
float sum_base(const test_base_t *p) {
float sum = 0.F;
for (size_t i = 0; i < p->data_size; i ) {
sum = p->data[i];
}
return sum;
}
void print_test(const test_base_t *p) {
printf("%"PRIu64" %zu { ", p->header, p->data_size);
if (p->data_size) {
printf("%g", p->data[0]);
for (size_t i = 1; i < p->data_size; i ) {
printf(" %g", p->data[i]);
}
}
printf("} sum=%g\n", sum_base(p));
}
int main() {
test_base_t *p1 = TEST_ARRAY(1){.data_size = 1, .data = {1}};
test_base_t *p2 = TEST_ARRAY(2){.data_size = 2, .data = {1, 2}};
print_test(p1);
print_test(p2);
print_test(TEST_ARRAY(3){.data_size = 3, .data = {1, 2, 3}});
print_test(TEST_ARRAY(4){.data_size = 4, .data = {1, 3, 5, 7}});
return 0;
}
CodePudding user response:
C 2018 6.7.2.1 3 says of a structure containing a flexible array member:
… such a structure (and any union containing, possibly recursively, a member that is such a structure) shall not be a member of a structure or an element of an array.
Thus, the test_t
type in the question violates this “shall” requirement, and C 2018 4 2 says that makes the behavior not defined by the C standard. A compiler could reject this code. If the compiler accepts it, the behavior of the program is not defined by the C standard.
As an example of what could go wrong (in that the C standard allows it), consider this code:
test_t test = { .base = { .data_size = 3, }, .data = { 1, 2, 3, }, };
printf("%g\n", test.base.data[0]);
Since test.base.data[0]
was never assigned a value through that expression, and the standard does not define test.data
to alias test.base.data
, the compiler may assume the value of test.base.data[0]
is uninitialized and hence unspecified, and this printf
may use any value of the float
type, even if test.base.data[0]
and test.data[0]
nominally refer to the same memory.
And in this code:
test_t test = { .base = { .data_size = 3, } };
for (int i = 0; i < 4; i)
test.base.data[i] = i 1;
test_t copy = test;
The compiler may assume that, since test.data
was never initialized, it does not need to be copied to copy
when initializing it from test
.