user: 这段汇编代码用C语言表示是什么样的？

```asm
<decode_base64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  shl    $0x2,%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movl   $0x0,-0x30(%rbp)
  jmp    L13
L23:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <ctoi@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x30(%rbp)
L13:
  mov    -0x30(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L23
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L29
L35:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x12,%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0xc,%eax
  or     %eax,%edx
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x2(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  shl    $0x6,%eax
  or     %eax,%edx
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x3(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  or     %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x24(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%edx
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x4,-0x30(%rbp)
  addl   $0x3,-0x2c(%rbp)
L29:
  mov    -0x30(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L35
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x40(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void decode_base64(char *str_in, char **str_out) {
  int i, j, b;
  int len_i = (strlen(str_in));
  char *str_t = (char *)malloc((len_i * 4) / 3);
  char *str_in_cp = (char *)malloc(len_i);
  strcpy(str_in_cp, str_in);
  for (i = 0; i < len_i; i++)
    str_in_cp[i] = ctoi(str_in_cp[i]);
  for (i = 0, j = 0; i < len_i; i += 4, j += 3) {
    b = (str_in_cp[i] << 18) | (str_in_cp[i + 1] << 12) | (str_in_cp[i + 2] << 6) | str_in_cp[i + 3];
    str_t[j] = b >> 16 & 0xff;
    str_t[j + 1] = b >> 8 & 0xff;
    str_t[j + 2] = b & 0xff;
  }
  free(str_in_cp);
  *str_out = str_t;
}
```