C++ : why bool is 8 bits long?

Because every C++ data type must be addressable.

How would you create a pointer to a single bit? You can't. But you can create a pointer to a byte. So a boolean in C++ is typically byte-sized. (It may be larger as well. That's up to the implementation. The main thing is that it must be addressable, so no C++ datatype can be smaller than a byte)

The answers about 8-bits being the smallest amount of memory that is addressable are correct. However, some languages can use 1-bit for booleans, in a way. I seem to remember Pascal implementing sets as bit strings. That is, for the following set:

{1, 2, 5, 7}

You might have this in memory:

01100101

You can, of course, do something similar in C / C++ if you want. (If you're keeping track of a bunch of booleans, it could make sense, but it really depends on the situation.)

A boolean type normally follows the smallest unit of addressable memory of the target machine (i.e. usually the 8bits byte).

Access to memory is always in "chunks" (multiple of words, this is for efficiency at the hardware level, bus transactions): a boolean bit cannot be addressed "alone" in most CPU systems. Of course, once the data is contained in a register, there are often specialized instructions to manipulate bits independently.

For this reason, it is quite common to use techniques of "bit packing" in order to increase efficiency in using "boolean" base data types. A technique such as enum (in C) with power of 2 coding is a good example. The same sort of trick is found in most languages.

Updated: Thanks to a excellent discussion, it was brought to my attention that sizeof(char)==1 by definition in C++. Hence, addressing of a "boolean" data type is pretty tied to the smallest unit of addressable memory (reinforces my point).

Memory is byte addressable. You cannot address a single bit, without shifting or masking the byte read from memory. I would imagine this is a very large reason.