user: 这段汇编代码用C语言表示是什么样的？

```asm
<alpha2_encode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L37
L59:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  and    $0xf,%eax
  mov    %al,-0x1e(%rbp)
  movzbl -0x1e(%rbp),%eax
  mov    %al,-0x1d(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  mov    %al,-0x1c(%rbp)
  movzbl -0x1c(%rbp),%eax
  mov    %al,-0x1b(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L29
L60:
  mov    -0x10(%rbp),%eax
  cltq
  lea    D6(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  movzbl -0x1d(%rbp),%eax
  cmp    %eax,%edx
  je     L79
  addl   $0x1,-0x10(%rbp)
L29:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x10(%rbp)
  jl     L60
  jmp    L71
L79:
  nop
L71:
  mov    -0x10(%rbp),%eax
  cltq
  lea    D6(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  sar    $0x4,%al
  mov    %al,-0x1b(%rbp)
  movzbl -0x1b(%rbp),%eax
  xor    -0x1d(%rbp),%al
  mov    %al,-0x1a(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L57
L38:
  mov    -0xc(%rbp),%eax
  cltq
  lea    D6(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  movzbl -0x1a(%rbp),%eax
  cmp    %eax,%edx
  je     L77
  addl   $0x1,-0xc(%rbp)
L57:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L38
  jmp    L99
L77:
  nop
L99:
  mov    -0xc(%rbp),%eax
  cltq
  lea    D6(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  sar    $0x4,%al
  mov    %al,-0x19(%rbp)
  movzbl -0x19(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl -0x1a(%rbp),%edx
  add    %ecx,%edx
  mov    %dl,(%rax)
  movzbl -0x1b(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl -0x1d(%rbp),%edx
  add    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x18(%rbp)
  addl   $0x2,-0x14(%rbp)
L37:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L59
  mov    -0x2c(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
data_type: string

tool: D6: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ

assistant: ```c
unsigned char *alpha2_encode(unsigned char *data, int len) {
  int i, j, m, n;
  unsigned char ah, al, A, B, C, D;
  unsigned char *newshellcode = malloc(2 * len + 1);
  for (i = 0, j = 0; i < len; i++, j += 2) {
    A = al = data[i] & 0x0f;
    B = ah = (data[i] & 0xf0) >> 4;
    for (m = 0; m < strlen("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"); m++) {
      if (("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"[m] & 0x0F) == A)
        break;
    }
    B = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"[m] >> 4;
    C = B ^ A;
    for (n = 0; n < strlen("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"); n++) {
      if (("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"[n] & 0x0F) == C)
        break;
    }
    D = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"[n] >> 4;
    newshellcode[j] = (D << 4) + C;
    newshellcode[j + 1] = (B << 4) + A;
  }
  newshellcode[2 * len] = '\x0';
  return newshellcode;
}
```