4 ms·

Talking theoretically, since size_t is defined as: > size_t can store the maximum size of a theoretically possible object of any type (including array). In a

Talking theoretically, since size_t is defined as: > size_t can store the maximum size of a theoretically possible object of any type (including array). In a proper theoretical turing machine, size_t would allow you to create and address arbitrarily large arrays too.

camel-cdr 4y ago
`sizeof(size_t)` and `size_t x; sizeof x` must be constant expressions. It is possible to create a c implementation where the size of a pointer is arbitrarily large, but it can't grow/isn't infinite. For any give C implementation, it is thus trivial to theoretically solve the halting problem (assuming no stdio.h stuff is used).
- Phrodo_00 4y ago
  Speaking in so much the abstract that it becomes absurd (which also can be used to prove your point, but bear with me here), since `size_t` is required to hold the size of any object, and a turing machine has infinite memory, then size_t would also need to be infinitely big, and `sizeof(size_t)` would be a constant returning infinity. So, I don't think it having to be a constant is a problem as much as having to deal with infinitely big numbers inside of infinite memory (which may or may not be a contradiction, depending on axioms used to define a turing machine. I still need to work my way through annotated turing). Of course, things do get a lot simpler and grounded using IO functions like you said earlier.
  - camel-cdr 4y ago
    I don't think infinitely large integer constant area construct that makes sense. As far as I'm aware, there isn't an instance of an infinitely large integer, even in mathematics, there are finite integers and there is the concept of infinity. (And size_t is defined as an unsigned integer type)
    - GabrielTFS 4y ago
      I'd argue you could use non-standard integers that are unable to be reached in a finite amount of steps but aren't infinity and are still "integers".
      - camel-cdr 4y ago
        > unable to be reached in a finite amount of steps but aren't infinity I don't see how such numbers can exit/be used/defined in any meaningful way. Do you have an example of such a number?

camel-cdr 4y ago

`sizeof(size_t)` and `size_t x; sizeof x` must be constant expressions. It is possible to create a c implementation where the size of a pointer is arbitrarily large, but it can't grow/isn't infinite. For any give C implementation, it is thus trivial to theoretically solve the halting problem (assuming no stdio.h stuff is used).

Phrodo_00 4y ago
Speaking in so much the abstract that it becomes absurd (which also can be used to prove your point, but bear with me here), since `size_t` is required to hold the size of any object, and a turing machine has infinite memory, then size_t would also need to be infinitely big, and `sizeof(size_t)` would be a constant returning infinity. So, I don't think it having to be a constant is a problem as much as having to deal with infinitely big numbers inside of infinite memory (which may or may not be a contradiction, depending on axioms used to define a turing machine. I still need to work my way through annotated turing). Of course, things do get a lot simpler and grounded using IO functions like you said earlier.
- camel-cdr 4y ago
  I don't think infinitely large integer constant area construct that makes sense. As far as I'm aware, there isn't an instance of an infinitely large integer, even in mathematics, there are finite integers and there is the concept of infinity. (And size_t is defined as an unsigned integer type)
  - GabrielTFS 4y ago
    I'd argue you could use non-standard integers that are unable to be reached in a finite amount of steps but aren't infinity and are still "integers".
    - camel-cdr 4y ago
      > unable to be reached in a finite amount of steps but aren't infinity I don't see how such numbers can exit/be used/defined in any meaningful way. Do you have an example of such a number?