CODEWITHSUNDEEP
memory-managementdata-structuresrust
o_o_o--
o_o_o--

Reputation: 1025

Why does Nil increase one enum size but not another? How is memory allocated for Rust enums?

If I define the following enums, Nil does not increase the size of the enum:

  use std::mem::size_of;

  enum Foo {
    Cons(~char)
  }

  enum Bar {
    Cons(~char),
    Nil
  }

  println!("{}", size_of::<Foo>());
  println!("{}", size_of::<Bar>());

  // -> 4
  // -> 4

On the other hand: 

  enum Foo {
    Cons(char)
  }

  enum Foo {
    Cons(char),
    Nil
  }

Yields:

  // -> 4
  // -> 8

What is happening when I define an enum? How is memory being allocated for these structures?

Upvotes: 5

Views: 762

Answers (1)

Chris Morgan
Chris Morgan

Reputation: 90832

A naive approach to enums is to allocate enough space for the contents of its largest variant, plus a descriminant. This is a standard tagged union.

Rust is a little cleverer than this. (It could be a lot cleverer, but it is not at present.) It knows that given a ~T, there is at least one value that that memory location cannot be: zero. And so in a case like your enum { Cons(~T), Nil }, it is able to optimise it down to one word, with any non-zero value in memory meaning Cons(~T) and a zero value in memory meaning Nil.

When you deal with char, that optimisation cannot occur: zero is a valid codepoint. As it happens, char is defined as being a Unicode code-point, so it would actually be possible to optimise the variant into that space, there being plenty of spare bits at the end (Unicode character only needs 21 bits, so in a 32-bit space we have eleven spare bits). This is a demonstration of the fact that Rust's enum discriminant optimisation is not especially clever at present.

Upvotes: 11

Related Questions