user: 这段汇编代码用C语言表示是什么样的：

```asm
<BaseEncode>:
  endbr64
  push   %r14
  mov    $0xaaaaaaab,%r11d
  mov    %rdi,%r10
  mov    %rcx,%r8
  push   %r13
  mov    %rdx,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  imul   %rbx,%r11
  mov    %rbx,%rbp
  shr    $0x21,%r11
  cmp    $0x2,%ebx
  jbe    L9
  mov    $0x1,%r13d
  mov    %r10,%rcx
  xor    %edx,%edx
  xor    %r9d,%r9d
  mov    $0x2,%r12d
  sub    %r10d,%r13d
  sub    %r10d,%r12d
  nopl   0x0(%rax,%rax,1)
L24:
  movzbl (%rcx),%eax
  add    $0x1,%r9d
  shr    $0x2,%al
  movzbl %al,%eax
  movzbl (%r8,%rax,1),%esi
  mov    %edx,%eax
  mov    %sil,(%rdi,%rax,1)
  lea    0x0(%r13,%rcx,1),%esi
  movzbl (%rcx),%r14d
  add    %r10,%rsi
  movzbl (%rsi),%eax
  shl    $0x6,%r14d
  shr    $0x2,%al
  or     %r14d,%eax
  shr    $0x2,%al
  movzbl %al,%eax
  movzbl (%r8,%rax,1),%r14d
  lea    0x1(%rdx),%eax
  mov    %r14b,(%rdi,%rax,1)
  lea    (%r12,%rcx,1),%r14d
  movzbl (%rsi),%esi
  add    $0x3,%rcx
  add    %r10,%r14
  movzbl (%r14),%eax
  shl    $0x4,%esi
  shr    $0x4,%al
  or     %esi,%eax
  shr    $0x2,%al
  movzbl %al,%eax
  movzbl (%r8,%rax,1),%esi
  lea    0x2(%rdx),%eax
  mov    %sil,(%rdi,%rax,1)
  movzbl (%r14),%eax
  and    $0x3f,%eax
  movzbl (%r8,%rax,1),%esi
  lea    0x3(%rdx),%eax
  add    $0x4,%edx
  mov    %sil,(%rdi,%rax,1)
  cmp    %r9d,%r11d
  ja     L24
  shl    $0x2,%r11d
  mov    $0x4,%eax
  cmp    $0x2,%ebp
  cmova  %r11d,%eax
L42:
  mov    $0xaaaaaaab,%edx
  imul   %rdx,%rbx
  shr    $0x21,%rbx
  lea    (%rbx,%rbx,2),%edx
  mov    %ebp,%ebx
  sub    %edx,%ebx
  cmp    $0x1,%ebx
  je     L25
  cmp    $0x2,%ebx
  jne    L39
  lea    -0x2(%rbp),%ecx
  add    %r10,%rcx
  movzbl (%rcx),%edx
  shr    $0x2,%dl
  movzbl %dl,%edx
  movzbl (%r8,%rdx,1),%esi
  mov    %eax,%edx
  mov    %sil,(%rdi,%rdx,1)
  lea    -0x1(%rbp),%edx
  movzbl (%rcx),%ecx
  add    %rdx,%r10
  movzbl (%r10),%edx
  shl    $0x6,%ecx
  shr    $0x2,%dl
  or     %ecx,%edx
  shr    $0x2,%dl
  movzbl %dl,%edx
  movzbl (%r8,%rdx,1),%ecx
  lea    0x1(%rax),%edx
  mov    %cl,(%rdi,%rdx,1)
  movzbl (%r10),%edx
  shl    $0x4,%edx
  shr    $0x2,%dl
  movzbl %dl,%edx
  movzbl (%r8,%rdx,1),%ecx
  lea    0x2(%rax),%edx
  add    $0x3,%eax
  mov    %cl,(%rdi,%rdx,1)
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  lea    -0x1(%rbp),%ecx
  pop    %rbx
  pop    %rbp
  add    %r10,%rcx
  pop    %r12
  pop    %r13
  movzbl (%rcx),%edx
  pop    %r14
  shr    $0x2,%dl
  movzbl %dl,%edx
  movzbl (%r8,%rdx,1),%esi
  mov    %eax,%edx
  mov    %sil,(%rdi,%rdx,1)
  movzbl (%rcx),%edx
  shl    $0x6,%edx
  shr    $0x2,%dl
  movzbl %dl,%edx
  movzbl (%r8,%rdx,1),%ecx
  lea    0x1(%rax),%edx
  add    $0x2,%eax
  mov    %cl,(%rdi,%rdx,1)
  ret
  nopl   (%rax)
L9:
  xor    %eax,%eax
  jmp    L42
```

assistant: ```c
unsigned int BaseEncode(unsigned char *hstr, unsigned int n, unsigned char *bstr, unsigned char *Bs64) {
  unsigned int i, m, c = 0, w = 0;
  unsigned char bas;
  m = (n / 3);
  for (i = 0; i < m; i++) {
    bas = (hstr[c] >> 2);
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[c] << 6) | (hstr[c + 1] >> 2);
    bas >>= 2;
    c++;
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[c] << 4) | (hstr[c + 1] >> 4);
    bas >>= 2;
    c++;
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[c] << 2);
    bas >>= 2;
    c++;
    bstr[w] = Bs64[bas];
    w++;
  }
  switch (n - (m * 3)) {
  case 1:
    bas = (hstr[n - 1] >> 2);
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[n - 1] << 6);
    bas >>= 2;
    bstr[w] = Bs64[bas];
    w++;
    break;
  case 2:
    bas = (hstr[n - 2] >> 2);
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[n - 2] << 6) | (hstr[n - 1] >> 2);
    bas >>= 2;
    bstr[w] = Bs64[bas];
    w++;
    bas = (hstr[n - 1] << 4);
    bas >>= 2;
    bstr[w] = Bs64[bas];
    w++;
    break;
  }
  return w;
}
```