What you are doing is undefined behavior.
The compiler is allowed to assume that you will never use more than sizeof int64_t for the variable member int64_t c. So if you try to write more than sizeof int64_t(aka sizeof c) on c, you will have an out-of-bounds problem in your code. This is the case because sizeof "aaaaaaaa" > sizeof int64_t.
The point is, even if you allocate the correct memory size using malloc(), the compiler is allowed to assume you will never use more than sizeof int64_t in your strcpy() or memcpy() call. Because you send the address of c (aka int64_t c).
TL;DR: You are trying to copy 9 bytes to a type consisting of 8 bytes (we suppose that a byte is an octet). (From @Kcvin)
If you want something similar use flexible array members from C99:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
size_t size;
char str[];
} string;
int main(void) {
char str[] = "aaaaaaaa";
size_t len_str = strlen(str);
string *p = malloc(sizeof *p + len_str + 1);
if (!p) {
return 1;
}
p->size = len_str;
strcpy(p->str, str);
puts(p->str);
strncpy(p->str, str, len_str + 1);
puts(p->str);
memcpy(p->str, str, len_str + 1);
puts(p->str);
free(p);
}
Note: For standard quote please refer to this answer.