user: 这段汇编代码用C语言表示是什么样的?

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%rsi
  mov    -0x40(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  add    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %eax,%edx
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  imul   %edx,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x40(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <SHA384@plt>
  movl   $0x1,-0x30(%rbp)
  jmp    L7
L2:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  imul   %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  mov    -0x30(%rbp),%ecx
  sub    $0x1,%ecx
  movslq %ecx,%rsi
  mov    D0(%rip),%rcx
  mov    (%rcx),%rcx
  imul   %rcx,%rsi
  mov    -0x28(%rbp),%rcx
  add    %rsi,%rcx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <SHA384@plt>
  addl   $0x1,-0x30(%rbp)
L7:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L2
  mov    -0x40(%rbp),%rax
  shl    $0x2,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  add    %rax,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  shr    %rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x40(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x40(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x40(%rbp),%rax
  shr    %rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x40(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x28(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned char *hash(const unsigned char *str, size_t len) {
  int j, i = (len / SHA384_DIGEST_LENGTH) + 1;
  unsigned char *digest = (unsigned char *)calloc(i * SHA384_DIGEST_LENGTH, sizeof(char));
  SHA384(str, len, digest);
  for (j = 1; j < i; ++j) {
    SHA384(digest + (SHA384_DIGEST_LENGTH * (j - 1)), SHA384_DIGEST_LENGTH, digest + (SHA384_DIGEST_LENGTH * j));
  }
  void *p = malloc(len * 4);
  void *x = malloc(len * 2);
  void *q = malloc(len / 2);
  void *r = malloc(len);
  memcpy(((char *)p) + len, digest, len);
  memcpy(((char *)x), digest, len);
  memcpy(q, digest, len / 2);
  memcpy(r, digest, len);
  free(p);
  free(x);
  free(q);
  free(r);
  return digest;
}
```