对 char 数组结构成员进行类型双关

Question

Consider the following code:

typedef struct { char byte; } byte_t;
typedef struct { char bytes[10]; } blob_t;

int f(void) {
  blob_t a = {0};
  *(byte_t *)a.bytes = (byte_t){10};
  return a.bytes[0];
}

Does this give aliasing problems in the return statement? You do have that a.bytes dereferences a type that does not alias the assignment in patch , but on the other hand, the [0] part dereferences a type that does alias.

I can construct a slightly larger example where gcc -O1 -fstrict-aliasing does make the function return 0, and I'd like to know if this is a gcc bug, and if not, what I can do to avoid this problem (in my real-life example, the assignment happens in a separate function so that both functions look really innocent in isolation).

Here is a longer more complete example for testing:

#include <stdio.h>

typedef struct { char byte; } byte_t;
typedef struct { char bytes[10]; } blob_t;

static char *find(char *buf) {
    for (int i = 0; i < 1; i++) { if (buf[0] == 0) { return buf; }}
    return 0;
}

void patch(char *b) { 
    *(byte_t *) b = (byte_t) {10}; 
}

int main(void) {
    blob_t a = {0};
    char *b = find(a.bytes);
    if (b) {
        patch(b);
    }
    printf("%d\n", a.bytes[0]);
}

Building with gcc -O1 -fstrict-aliasing produces 0

Answer 1

The main issue here is that those two structs are not compatible types. And so there can be various problems with alignment and padding.

That issue aside, the standard 6.5/7 only allows for this (the "strict aliasing rule"):

An object shall have its stored value accessed only by an lvalue expression that has one of the following types:

• a type compatible with the effective type of the object,
  ...
• an aggregate or union type that includes one of the aforementioned types among its members

Looking at *(byte_t *)a.bytes , then a.bytes has the effective type char[10] . Each individual member of that array has in turn the effective type char . You de-reference that with byte_t , which is not a compatible struct type nor does it have a char[10] among its members. It does have char though.

The standard is not exactly clear how to treat an object which effective type is an array. If you read the above part strictly, then your code does indeed violate strict aliasing, because you access a char[10] through a struct which doesn't have a char[10] member. I'd also be a bit concerned about the compiler padding either struct to meet alignment.

Generally, I'd simply advise against doing fishy things like this. If you need type punning, then use a union. And if you wish to use raw binary data, then use uint8_t instead of the potentially signed & non-portable char .

Answer 2

The error is in *(byte_t *)a.bytes = (byte_t){10}; . The C spec has a special rule about character types (6.5§7), but that rule only applies when using character type to access any other type, not when using any type to access a character.

Answer 3

According to the Standard, the syntax array[index] is shorthand for *((array)+(index)) . Thus, p->array[index] is equivalent to *((p->array) + (index)) , which uses the address of p to compute the address of p->array , and then without regard for p 's type, adds index (scaled by the size of the array-element type), and then dereferences the resulting pointer to yield an lvalue of the array-element type. Nothing in the wording of the Standard would imply that an access via the resulting lvalue is an access to an lvalue of the underlying structure type. Thus, if the struct member is an array of character type, the constraints of N1570 6.5p7 would allow an lvalue of that form to access storage of any type.

The maintainers of some compilers such as gcc, however, appear to view the laxity of the Standard there as a defect. This can be demonstrated via the code:

struct s1 { char x[10]; };
struct s2 { char x[10]; };
union s1s2 { struct s1 v1; struct s2 v2; } u;

int read_s1_x(struct s1 *p, int i)
{
    return p->x[i];
}
void set_s2_x(struct s2 *p, int i, int value)
{
    p->x[i] = value;
}
__attribute__((noinline))
int test(void *p, int i)
{
    if (read_s1_x(p, 0))
        set_s2_x(p, i, 2);
    return read_s1_x(p, 0);
}
#include <stdio.h>
int main(void)
{
    u.v2.x[0] = 1;
    int result = test(&u, 0);
    printf("Result = %d / %d", result, u.v2.x[0]);
}

The code abides the constraints in N1570 6.5p7 because it all accesses to any portion of u are performed using lvalues of character type. Nonetheless, the code generated by gcc will not allow for the possibility that the storage accessed by (*(struct s1))->x[0] might also be accessed by (*(struct s2))->x[i] despite the fact that both accesses use lvalues of character type.