I think you must be referring to the example on varint encoding here:
https://developers.google.com/protocol-buffers/docs/encoding I believe the
example is correct; it's just that the ++ represents a concatenation
instead of an addition. Each byte has 7 bits for storing the number (since
the remaining bit is used to indicate whether there are more bytes), and so
we are concatenating them to get the full 14 bits of the actual number.
Note also that the five leading zeros are omitted.

On Thu, Sep 22, 2016 at 1:35 AM, <yanxi9...@gmail.com> wrote:

> 1. 000 0010  010 1100
> 2. →  000 0010 ++ 010 1100
> 3. →  100101100
> 4. →  256 + 32 + 8 + 4 = 300
>
> i have a doubt
>
> 000 0010 ++ 010 1100 = 1 0010 1100, should not 010 1110 ?
>
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