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"Functionally equivalent" to, not identical to. I'm talking about the module system. You could literally replace any instance of: mod mod_name; With:

"Functionally equivalent" to, not identical to. I'm talking about the module system. You could literally replace any instance of: mod mod_name; With: #include "path/to/mod_name" and then run it through the C preprocessor and it'd work exactly the same. The discussion was held on IRC.

steveklabnik 12y ago
This is absolutely not true. Rust has no textual substitution like the C preprocessor does, and includes a module system, which has namespacing.
- Sir_Cmpwn 12y ago
  I put together a small demonstration for you: https://github.com/SirCmpwn/rust-poc https://github.com/SirCmpwn/rust-poc
  - steveklabnik 12y ago
    That is one, very, very tiny aspect of the C preprocessor. And https://github.com/SirCmpwn/rust-poc/blob/master/includes.rs#L1 https://github.com/SirCmpwn/rust-poc/blob/master/includes.rs... is significant: the code included from another file is under a namespace, and isn't going to clash like randomly throwing #includes around will, and you don't need to worry about things like double inclusion... In other words, I think this is a very superficial comparison, and I'm not even sure why it upsets you so. Rust won't be a perfect language for everyone, and that's okay.
    - Sir_Cmpwn 12y ago
      Well, you could move the mod { } to the included file to accomplish the same result (like the `namespace` keyword in C#), but that'd break the example I'm trying to give. I don't really want to rehash this argument on Hacker News, I'm just providing context. There are numerous flaws in the module system that derive from this basic point.
phaylon 12y ago
I'm not really a C guy, but isn't #include about using a module in the current scope? Because Rust's mod is for project structure definition.
- Sir_Cmpwn 12y ago
  I can't really effectively explain how all of this works without providing a lot of background on systems languages in general, especially background on how linking works. Sorry.
- dbaupp 12y ago
  #include in C is just textual substitution of the contents of the specified file in place of the #include statement, there is not a concept of namespace in C.

steveklabnik 12y ago

This is absolutely not true. Rust has no textual substitution like the C preprocessor does, and includes a module system, which has namespacing.

Sir_Cmpwn 12y ago
I put together a small demonstration for you: https://github.com/SirCmpwn/rust-poc https://github.com/SirCmpwn/rust-poc
- steveklabnik 12y ago
  That is one, very, very tiny aspect of the C preprocessor. And https://github.com/SirCmpwn/rust-poc/blob/master/includes.rs#L1 https://github.com/SirCmpwn/rust-poc/blob/master/includes.rs... is significant: the code included from another file is under a namespace, and isn't going to clash like randomly throwing #includes around will, and you don't need to worry about things like double inclusion... In other words, I think this is a very superficial comparison, and I'm not even sure why it upsets you so. Rust won't be a perfect language for everyone, and that's okay.
  - Sir_Cmpwn 12y ago
    Well, you could move the mod { } to the included file to accomplish the same result (like the `namespace` keyword in C#), but that'd break the example I'm trying to give. I don't really want to rehash this argument on Hacker News, I'm just providing context. There are numerous flaws in the module system that derive from this basic point.

phaylon 12y ago

I'm not really a C guy, but isn't #include about using a module in the current scope? Because Rust's mod is for project structure definition.

Sir_Cmpwn 12y ago
I can't really effectively explain how all of this works without providing a lot of background on systems languages in general, especially background on how linking works. Sorry.
dbaupp 12y ago
#include in C is just textual substitution of the contents of the specified file in place of the #include statement, there is not a concept of namespace in C.