RemainingSeconds is the same for all codes in same time step. #59
Description
Why does generating a new code within the same time step but with a different key not have a new RemainingSeconds?
If I call GetTotp two times with a step of 9 minutes, the second call has a RemainingSeconds of the first call.
GetTotp takes a byte[] and count, 1, 2, 3, etc. the byte[] key is a hashed salt and count which generates a new code. I found that I had to make the key more unique than the previous call to GetTotp because calling GetTotp again (like when user wants a new key) within the 9 minute step just returned the same code.
Example. time step is 540 seconds.
GetTotp(salt, 1);
"309960"
RemainingSeconds 539
wait some seconds and call again...
GetTotp(salt, 2);
"718933"
RemainingSeconds 460
I expected RemainingSeconds to be closer to 539
internal string GetTotp(byte[] salt, int count)
{
string saltedKey = HashSomeStringWithSalt(salt, count.ToString());
byte[] key = Convert.FromBase64String(saltedKey);
var totp = new Totp(key, step: 540, mode: OtpHashMode.Sha256, totpSize: 6);
var dateTimeNow = DateTime.UtcNow;
var totpCode = totp.ComputeTotp(dateTimeNow);
return totpCode;
}
internal int GetOTPRemainingSeconds(byte[] salt, int count)
{
string saltedKey = HashSomeStringWithSalt(salt, count.ToString());
byte[] key = Convert.FromBase64String(key);
var totp = new Totp(key, step: 540, mode: OtpHashMode.Sha256, totpSize: 6);
return totp.RemainingSeconds();
}