user: 请从给定的汇编指令推导出C源代码：

```asm
<hlp_file_randmod>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    $0x1,%edi
  push   %r13
  mov    %r8,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rcx,(%rsp)
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %rax,%rax
  je     L1
  mov    D34(%rip),%rax
  mov    %r14,%rdi
  mov    (%rax),%esi
  call   <open@plt>
  mov    %eax,0x14(%rsp)
  test   %eax,%eax
  js     L89
  mov    D20(%rip),%rax
  mov    0x14(%rsp),%edi
  xor    %esi,%esi
  mov    (%rax),%edx
  call   <lseek@plt>
  mov    %eax,0x10(%rsp)
  test   %eax,%eax
  js     L20
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    $0x1,%r15d
  call   <sranddev@plt>
  mov    %r12d,%eax
  mov    %rax,0x18(%rsp)
  test   %r12d,%r12d
  je     L46
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%r14d
  lea    -0x1(%rbx),%r12d
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idivl  0x10(%rsp)
  mov    %edx,0x0(%rbp,%rbx,4)
  cmp    %r14d,%r12d
  ja     L58
  mov    %r12d,%eax
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L56:
  mov    %eax,%esi
  cmp    0x0(%rbp,%rsi,4),%edx
  cmove  %r15d,%ecx
  sub    $0x1,%eax
  cmp    %r14d,%eax
  jbe    L56
  test   %ecx,%ecx
  jne    L40
  cs nopw 0x0(%rax,%rax,1)
L58:
  mov    0x8(%rsp),%rsi
  mov    0x14(%rsp),%edi
  mov    %edx,%ecx
  mov    $0x1,%edx
  call   <pread@plt>
  cmp    $0x1,%eax
  jne    L0
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rdi
  movzbl (%rax),%eax
  mov    %al,(%rdi,%rbx,1)
  mov    %al,0x0(%r13)
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  call   <rand@plt>
  mov    (%rsp),%rdi
  cltd
  shr    $0x18,%edx
  add    %edx,%eax
  movzbl %al,%eax
  sub    %edx,%eax
  mov    %al,0x0(%r13)
  cmp    (%rdi,%rbx,1),%al
  je     L3
  mov    0x0(%rbp,%rbx,4),%ecx
  mov    0x14(%rsp),%edi
  mov    $0x1,%edx
  mov    %r13,%rsi
  call   <pwrite@plt>
  cmp    $0x1,%eax
  jne    L0
  add    $0x1,%rbx
  add    $0x1,%r13
  cmp    %rbx,0x18(%rsp)
  jne    L8
L46:
  mov    0x14(%rsp),%edi
  call   <close@plt>
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  mov    0x14(%rsp),%edi
  call   <close@plt>
L89:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
L1:
  mov    D23(%rip),%rax
  mov    (%rax),%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    0x14(%rsp),%edi
  call   <close@plt>
  mov    D23(%rip),%rax
  mov    (%rax),%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int hlp_file_randmod(const char *path, unsigned int mod_amount, off_t *positions, unsigned char *orig_chars, unsigned char *mod_chars) {
  off_t filesize;
  int fd;
  unsigned int i, j, flag;
  void *buf = NULL;
  buf = malloc(1);
  if (NULL == buf)
    return errno;
  fd = open(path, O_RDWR);
  if (fd < 0) {
    free(buf);
    return errno;
  }
  filesize = lseek(fd, 0, SEEK_END);
  if (filesize < 0) {
    free(buf);
    close(fd);
    return errno;
  }
  sranddev();
  for (i = 0; i < mod_amount; i++) {
    flag = 1;
    while (flag) {
      positions[i] = (rand() % filesize);
      flag = 0;
      for (j = i - 1; j <= i; j--)
        if (positions[j] == positions[i])
          flag = 1;
    }
    if (1 != pread(fd, buf, 1, positions[i])) {
      close(fd);
      free(buf);
      return errno;
    }
    orig_chars[i] = (*(unsigned char *)buf);
    mod_chars[i] = orig_chars[i];
    while (mod_chars[i] == orig_chars[i])
      mod_chars[i] = (unsigned char)(rand() % 256);
    if (1 != pwrite(fd, &(mod_chars[i]), 1, positions[i])) {
      close(fd);
      free(buf);
      return errno;
    }
  }
  close(fd);
  free(buf);
  return 0;
}
```