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Understanding alignment in g++

Hi, consider this simple program:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <type_traits>
using namespace std;

enum { sz = 8 };

int main() {
  cout << alignment_of<int>::value << endl;

  char a[sz];
  cout << (void*)a << endl;
  bzero(a, sz);

  *((int*)(a+1)) = -1;
  for (int i = 0; i < sz; ++i) printf("%02hhx ", a[i]);
  cout << endl;

  return 0;

Here's the stdout on x86_64:

00 ff ff ff ff 00 00 00

I expected a cast-align warning or something. How did gcc make the unaligned write work? It didn't generate any inefficient code or anything; here's an excert from objdump -DE:

  *((int*)(a+1)) = -1;
  400a4a:       48 8d 45 f0             lea    -0x10(%rbp),%rax
  400a4e:       48 83 c0 01             add    $0x1,%rax
  400a52:       c7 00 ff ff ff ff       movl   $0xffffffff,(%rax)

Is the alignment_of value incorrect? I also tried __alignof__ and got the same result.

Related: Is there an example of a type that actually has non-1 alignment on x86(_64)? What happens if I try to use it as shown above (unaligned) - does the compiler just produce inefficient code? Does the memory access fall into alignment (thus producing unexpected results)?

Thanks in advance for any answers.
Yang Zhang

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