Does an OCaml char represent exactly 8 bits?

The document says this:

Character values are represented as 8-bit integers between 0 and 255. Character codes between 0 and 127 are interpreted following the ASCII standard. The current implementation interprets character codes between 128 and 255 following the ISO 8859-1 standard.

So yes, an OCaml char represents exactly 8 bits.

The documentation for base values is here: OCaml Manual, Chapter 9.2. Values.

Update

It might be worth noting that although a char value in OCaml can take on only values from 0 to 255, in the mainline OCaml version (from INRIA) the actual space occupied in memory by a char value is the same as for int. On a 32-bit implementation this will be 32 bits and on a 64-bit implementation it will be 64 bits. So (for example) a char array is not a space-efficient way to store more than a few chars. You can use string or bytes to get compact storage of char values (as 8 bits each).

The documentation for representation of OCaml values is here: OCaml Manual, Chapter 20.3, Representation of OCaml Data Types.

The representation of the char type could be different depending on the implementation of the OCaml language and runtime. While all chars shall fit into 8 bits, an implementation may use a bigger type to represent it. The Char abstraction guarantees that it is impossible to create a character that uses more than 8 bits. And even though the INRIA implementation of OCaml represents Char.t the same as Int.t, it still relies on the assumption that char will fit into 8 bits. For example, a bigarray of n chars will take n bytes. And String.t will have a size in bytes proportional to the number of characters that comprise the string. Last but not least, various external (i.e., implemented in C) functions and the optimized compiler itself will assume that a character fits into 8 bits.