The docs say, "one bit is reserved for the OCaml runtime", so doesn't that mean that one of the bits (likely the high bit) are unavailable for the programmer's use?
I mean, I understand "reserved" to mean either "you can't depend upon it if you use it", or "it will break the runtime if you use it".
So the "one bit" you refer to is what makes the standard int 63 bits rather than 64. If you could do things with it it would indeed break the runtime- that's what tells it that you're working with an int rather than a pointer. But full, real, 64-bit integers are available, in the base language, same goes for 32.
I think you need to re-read some of the comments you are replying to. There is a 64 bit int type: https://ocaml.org/manual/5.3/api/Int64.html
You can use all 64 bits. There are also other int types, with different amounts of bits. For example, 32 bit: https://ocaml.org/manual/5.3/api/Int32.html
No one will stop you. You can use all the bits you want. Just use the specific int type you want.
ravi-delia explained that the fact is that an OCaml int is different than either Int32 or Int64 because an 'int' sacrifices one of its bits to the OCaml runtime. Int32 or Int64 are treated completely differently and are library defintions, bolted onto the OCaml runtime.
That is a runtime system not suitable for systems-level programming.
My C experience gave me a fundamental misunderstanding because there, an int is always derived from either an 32- or 64-bit int, depending on architecture.
OCaml is architected differently. I imagine the purpose was to keep the programs mostly working the same across processor architecture sizes.
I imagine this fundamental difference between OCaml's native int and these more specific Ints is why there are open issues in the libray that I"m sure the int does not.
Regardless, no one should be using OCaml for systems-level programming.
Thanks for helping me get to the heart of the issue.
Ok, running this by you one more time. There is a type called "int" in the language. This is a 63-bit signed integer on 64-bit machines, and a 31-bit integer on 32-bit machines. It is stored in 64 bits (or 32), but it's a 63-bit signed integer, because one of the bits is used in the runtime.
There is also a 64 bit integer, called "Int64". It has 64 bits, which is why I call it a 64-bit integer rather than a 63-bit integer. An "int" is a 63-bit integer, which is why I call it a 63-bit integer rather than a 64-bit integer.
So an int has nothing to do with an Int32 or Int64.
Thanks for your patient elucidation.
This means the semantics for Int32 and Int64 are COMPLETELY different than that of an int. My problem is that I come from the C world, where an int is simply derived from either a 32- or 64-bit integer, depending on the target architecture.
OCaml's runtime is not a system designed for systems-level programming.
Thanks again.
Now I know why the F# guys rewrote OCaml's fundamental int types from the get-go.
> Performance notice: values of type int64 occupy more memory space than values of type int
I just couldn't even imagine that a 64-bit int would require MORE memory than an int that is one bit less (or 33 bits less if on a 32-bit architecture).
It really makes absolutely no sense discussing OCaml as a possible systems-level programming language.
bruh, it's just saying single scalar Int64 types are boxed. This is totally normal thing that happens in garbage collected languages. There's no semantic loss.
OCaml does this 63-bit hack to make integers fast in the statistically common case where people don't count to 2^64 with them. The top bit is reserved to tell the GC whether it manages the lifetime of that value or not.
For interoperating with binary interfaces you can just say `open Int64` at the top of your file and get semantic compatibility. The largest industrial user of OCaml is quant finance shop that binds all kinds of kernel level drivers with it.
(and yes, 64-bit non-boxed array types exist as well if you're worried about the boxing overhead)