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