user: What would the C source look like for the assembly code provided here?

```asm
<MD5_Final>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x30(%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x80,(%rax)
  addq   $0x1,-0x18(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  sub    $0x8,%eax
  cltq
  cmp    %rax,-0x18(%rbp)
  jbe    L8
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movq   $0x0,-0x18(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <md5_block_data_order@plt>
L8:
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x8,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  cltq
  sub    $0x8,%rax
  add    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0xc(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x10(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  cltq
  neg    %rax
  add    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <md5_block_data_order@plt>
  mov    -0x30(%rbp),%rax
  movq   $0x0,(%rax)
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <OPENSSL_cleanse@plt>
  mov    -0x30(%rbp),%rax
  mov    0x10(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x18(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x20(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x28(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x28(%rbp)
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int MD5_Final(unsigned char *md, MD5_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (MD5_CBLOCK - 8)) {
    memset(p + n, 0, MD5_CBLOCK - n);
    n = 0;
    md5_block_data_order(c, p, 1);
  }
  memset(p + n, 0, MD5_CBLOCK - 8 - n);
  p += MD5_CBLOCK - 8;
  (void)(*((unsigned int *)(p)) = (c->Nl), (p) += 4, c->Nl);
  (void)(*((unsigned int *)(p)) = (c->Nh), (p) += 4, c->Nh);
  p -= MD5_CBLOCK;
  md5_block_data_order(c, p, 1);
  c->num = 0;
  OPENSSL_cleanse(p, MD5_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4, ll);
    ll = (c)->B;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4, ll);
    ll = (c)->C;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4, ll);
    ll = (c)->D;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4, ll);
  } while (0);
  return 1;
}
```