string pointer to byte conversion condition

Question

I am trying to understand the following snippet. I understand that the first check would check if it is even if it would be an int, but I don't understand it in this context.

Could someone explain to me what this does?

if (((byte)*password & 1) == 0) {
  buf = password + 1;
  len = (uint)((byte)*password >> 1);
}
else {
  len = *(uint *)(password + 4);
  buf = *(basic_string **)(password + 8);
}

bp = BIO_new_mem_buf(buf,len);

Answer 1

Based on this limited amount of psuedo-C:

struct password_buffer {
    byte len:7;
    byte use_pointer:1;
    union {
        byte buffer[128];
        struct {
            byte pad[3];
            uint len;
            char *buf;
        } pointer;
    };
};

This "password" should be probably be treated like a union. That low bit is not odd or even, it's a flag to decide if the password is a buffer or in a pointer.

When the least significant bit is unset in the first byte, then the password is a buffer with the zero index byte containing the length in the high 7 bits followed directly by the password.

When the least significant bit is set in the first byte, then the length is the first uint after this first byte (4-byte aligned) and password is stored in memory with a pointer.

This structure probably will default to the "buffer" when the string is less than 128 characters; otherwise, it will be in other memory with a pointer to the memory and a length value.

structs like this will not cleanly be represented by most decompilation tools. There's just not enough information in the assembly to do so.