PPT eavesdroper able to output $m_{0}$ and $m_{1}$ of different lengths I've read the following two questions and their answers:


Messages of different lengths and one-time computationally-secret
Why is a non fixed-length encryption scheme worse than a fixed-length one?


I am still not convinced of what is the actual strategy of the adversary $A$. I understand that in case that the encryption scheme cipher depends on the length of the message to encrypt then $A$ could output $m_{0}$, $m_{1}$ of different lengths and would be able to tell which one was encrypted by the length of $Enc(m_{b})$ (where $b\gets\{0,1\}$).

Now, who promises that $|Enc(m)|$ depends on $|m|$?
Isn't it possible that $|Enc(m_{0})|=|Enc(m_{1})|$?
Actually, every encryption algorithm has an associated message space which defines the maximum message size which can be encrypted and the ciphertext is always of size of the associated ciphertext space. If the message at hand is larger than the message space of the encryption algorithm, you have to encrypt the message "block by block" (where a block can also be a bit. For block ciphers there are various ways to realise this, e.g., CBC (see here), for stream ciphers viewing the message space as being a bit it should be obvious how the ciphertext size depends on the message size and for asymmetric schemes in practice this is typically not required since you use hybrid encryption and thus always only encrypt a single "block". 

Now for achieving the weakest reasonable security, i.e., indistinguishability under chosen plaintext attacks (IND-CPA) where i'm talking about public key encryption schemes here (actually the argument is the same for one-time computationally-secret as in your referenced question), an adversary providing two challenge messages $m_0$ and $m_1$ and given back the ciphertext $c_b$ has to figure out which one has been encrypted (i.e., determine whether $b=0$ or $b=1$) (here it does not really matter that the adversary can then afterwards ask an encryption oracle for encryptions of messages different from the challenge messages). Clearly, if one of the challenge messages is out of the message space and the other is larger, i.e., requires $&gt;1$ blocks, then it is trivial to distinguish the two ciphertexts.

Consequently, in the attack game one requires that if one message is larger, the second is padded to have the same length as the first to rule out this trivial attack. Then, if both messages are of the same length (and padded) both ciphertexts will be of the same length and then the adversary has to apply a non-trivial strategy to figure out which message has been encrypted as a challenge.

Edit: PrivK Game:

Ok, the PrivK game is defined as follows (I took the definition from this question):

An (efficient secret-key) encryption scheme $(Gen,Enc,Dec)$ is one-time computationally-secret if for any PPT adversary $\cal A$ it holds that $Pr[PrivK_{\cal A}^{eav}(n)=1]−1/2$ is negligible function, where $PrivK_{\cal A}^{eav}(n)$ denotes the output of the following experiment:

(a) The adversary $\cal A$ on input $1^n$ outputs a pair of messages $m_0,m_1$.

(b) Let $k\leftarrow Gen(1^n)$ and let $b\in \{0,1\}$ be chosen uniformly at random. Then a ciphertext $c\leftarrow Enc_k(m_b)$ is computed and given to $\cal A$.

(c) $\cal A$ on input $c$ outputs a bit $b'$.

(d) The output of the experiment is $1$ if $b'=b$ and $0$ otherwise.

Let us assume that we have a block cipher with block size $k$ bits and let the adversary now choose a message $m_0\in \{0,1\}^k$ and $m_1 \in \{0,1\}^{2k}$, i.e., message $m_0$ has one block and message $m_1$ has two blocks. Consequently, $c_0=Enc_k(m_0)$ will be of size $k$ bits and $c_1=Enc_k(m_1)$ will be of size $2k$ bits. Hence, the adversary will always output $1$ and thus $Pr[PrivK_{\cal A}^{eav}(n)=1]=1$ and $Pr[PrivK_{\cal A}^{eav}(n)=1]-1/2$ is $1/2$, which is non negligible. Consequently, we need to require that $|m_0|=|m_1|$ to rule out this strategy for $\cal A$.