encryption - Why do AES decryption methods take IVs as arguments -

I am still new to encryption normally, and I understand why an IV is required, but IV embedded Not in encrypted data? Thus, any decryption implementation should first be deducted from the data instead of the requirement of this supply?

I think that Ivy is embedded in data because some libraries / devices do not require it to be supplied for example JavaScript, for CryptoJS. CryptoJS.AES.decrypt ("U2FsdGVkX1 / l3HWODO9GX23rvF0KHmDR6z8XTpYYpe8 =", "password") toString (CryptoJS.enc.Utf8));

and openssl: echo "U2FsdGVkX1 ++ PGG + oWqZbIjccV1NiV2pc1QrQtw0wp4 =" | Openssl AES 256 CBC-D-Passed A-Pass:. Password

And that AES encrypts, really like any True Application

, like TrueCrypt is a library that looks like IV:

• mcrypt for PHP:
• In the SymmetricAlgorithm net:

  I'm having some confusion < P>

  No, the fourth can be equally derivative. The only reason to include IV with Ciphertext is when the key is reused for encryption and when IV can not be obtained. It is normal, but not standardized or it is necessary for prefix if it exists. But when you need IV, this does not have anyone with the D-Facto standard.

  In your OpenSSL example a random salt is used to generate a key from the password. In that case salt is different for each encrypt, so the key is also random. In that case there is no need for a random IV, although it can be obtained with salt and password (and will be for OpenSSL).

  There is a counter in the other protocol, and in that case, a single block encryption counter can be used to create one quarter for CBC mode encryption. As long as the counter resides on the sender / receiver, there is no need to send IV.

  And there are so many cryptograhic implementations that get it wrong and do not use a random IV for CBC mode. PHP mcrypt_encrypt What is the special thing in the sense that it does not use a fourth anywhere in the algorithm for ECB mode, which is then ignored as thouroughly ECB requires a fourth is.