Big Endian conventions or what? Let be $m \in \{0,1\}^{l}$ and $r \in \{0,1\}^{|N|-l-1}$. 
I have seen that $m||r$ (where || means the concatenation of the bits) can be seen like an integer, I think using a Big-Endian Convention, and this integer is $m + 2^{l}r$. 
The question is why?
What is the way to convert the bits-string $m||r$ to the integer $2^l r$.
I have to understand this because, it is useful to  solve some exercises.

Thank you.
If $m\|r$ is assimilated to the integer $m+2^{l}r$, with $m\in\{0,1\}^{l}$, then little-endian convention is being used.



More precisely: under (unsigned) little-endian convention, a bitstring of $k$ bits $b=b_0\|b_1\|\dots\|b_{k-1}$ where each $b_i$ is a single bit is assimilated to the (non-negative) integer $\tilde b\,=\,\displaystyle\sum_{i=0}^{k-1}b_i2^i$&nbsp;. It follows that if $m$ and $r$ are bitstrings, the bitstring $m\|r$ is assimilated to the integer $\widetilde{m\|r}\,=\,\tilde m+2^{|m|}\tilde r$ (where $|m|$ is the number of bits in bitstring $m$ ).

For example, if $m\,=\,\mathtt{'00110101'}$ and $r\,=\,\mathtt{'1010'}$&nbsp;, then $|m|=l=8$&nbsp;, bitstring $m$ is assimilated to the integer $\tilde m=2^2+2^3+2^5+2^7=172$&nbsp;, bitstring $r$ is assimilated to the integer $\tilde r=2^0+2^2=5$&nbsp;, bitstring $m\|r\,=\,\mathtt{'001101011010'}$ and is assimilated to the integer $\widetilde{m\|r}\,=\,2^2+2^3+2^5+2^7+2^8+2^{10}=1452$&nbsp;, and indeed $172+2^8\cdot5=1452$&nbsp;.



Under (unsigned) big-endian convention, $b=b_0\|b_1\|\dots\|b_{k-1}$ is assimilated to the integer $\hat b\,=\,\displaystyle\sum_{i=0}^{k-1}b_i2^{k-1-i}$&nbsp;. It follows that $m\|r$ is assimilated to the integer $\widehat{m\|r}\,=\,2^{|r|}\hat m+\hat r$&nbsp;.

For example, if $m\,=\,\mathtt{'00110101'}$ and $r\,=\,\mathtt{'1010'}$&nbsp;, then $|r|=4$&nbsp;, $\hat m=2^6+2^5+2^2+2^0=101$&nbsp;, $\hat r=2^3+2^1=10$&nbsp;, $m\|r\,=\,\mathtt{'001101011010'}$&nbsp;, $\widehat{m\|r}\,=\,2^{10}+2^9+2^6+2^4+2^3+2^1=1626$&nbsp;, and indeed $2^4\cdot101+10=1626$&nbsp;.



The notation $\tilde{}$ for little-endian and $\hat{}$ for little-endian is entirely made up for the purpose of that answer, and omitted in practice. Endianness is considered too mundane to be made explicit, but/thus is a cause of endless headaches, incompatibilities, frustrations, and conversion code.

It is possible to use one convention for bits in $m$ and $r$, and the formula of the other convention for $m\|r$&nbsp;, that is assimilate $m\|r$ to integer $\hat m+2^{|m|}\hat r$ or $2^{|r|}\tilde m+\tilde r$. That's mixed endianness, and surprisingly common, often with change of endianness at regular intervals, e.g. at the octet (8 bits) or/and word (e.g. 4 octet) level. In that case, the formula converting bits to integer grows hairy (the practitioner's hair looking at bits with a debugging tool are correspondingly pulled out).



Most humans (including cryptographers unless otherwise stated, perhaps obliquely by throwing $m+2^{l}r$ in) tend to use big-endian convention, for natural reasons: when we are told twenty-five predators are approaching, twenty goes before five because that gives the order of magnitude sooner. The same ordering is kept when writing 25.

Little-endian is often used for transmission of individual bits in bytes, historically because it eased interoperability between equipments using asynchronous communication and different character width. However network byte order, or network order for short, refers to big-endian convention at the upper level (if you look at how a 16-bit IP port number is physically transmitted on a network, e.g. with an oscilloscope across the contacts of the LED transmitting into optical fiber, a 10Mbit/s Ethernet twisted pair, or an old modem, you'll find that's with little-endian convention within an octet, and with the high-order octet first, that is big-endian at a higher level).

Intel's convention is essentially opposite: it uses little-endian byte order (the low-order octet of a word has the lowest address, and documentation shows it first whenever octets are separated from each others); but when documentation shows an octet, that's with big-endian convention (always for decimal and hexadecimal, often for binary; however bits remain numbered with 0 for the low-order bit; ISO standards often start at 1 to make things worse, but I digress).

Little-endian convention removes the need for buffer memory when adding integers represented by bitstrings, octet strings, or vectors of limbs made available sequentially or by increasing address (serial memories or CPUs favoring index increment over decrement; ironically, the later applies to the Motorola 68000 CPU, which use big-endian octet order). That makes little-endian popular for implementation of multiple-precision arithmetic, together with the fact that realloc() can be used to adjust storage without changing the value represented.

Big-endian and Little-endian refer to names of the two camps in a fictional war, which wanted to impose on the other side the opening of eggs by the big or little end; see Gargantua Gulliver's Travels for details. This should remind us that "more sensible xxxx endian" should be taken like eggs: with a grain of salt.



The two most common conventions for signed integers are nowadays:


Two's complement, where signed integer $x$ is represented by bitstring $b$ such that $-2^{|b|-1}\le x&lt;2^{|b|-1}$ and the unsigned integer $\tilde b$ or $\hat b$ is $x\bmod2^{|b|}$&nbsp;, lying in interval $[0\dots2^{|b|})$.
Sign and magnitude, where signed integer $x$ is represented by bitstring $b$ such that $-2^{|b|-1}&lt;x&lt;2^{|b|-1}$ and the unsigned integer $\tilde b$ or $\hat b$ is


$2^{|b|-1}-x$ when $x&lt;0$
$x$, otherwise



Under both sign conventions, one of the bit of $b$ (the first when combined with big-endain convention, the last with little-endinan) is the sign bit, with $1$ for negative.

Under two's-complement, the result of addition modulo $2^{|b|}$ can be obtained using the same rules for unsigned and signed integers (only overflow detection varies). This explains why two's-complement is common in modern CPUs.