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I'm really happy with Meson, as a lot of Wayland (the new display protocol for Linux, the "successor" of X11) apps seem to be built in C, so using meson is sup

I'm really happy with Meson, as a lot of Wayland (the new display protocol for Linux, the "successor" of X11) apps seem to be built in C, so using meson is super simple and I don't have to worry about tooling (I don't deal much with C/C++, so let me make my change and run away please). Rust is the same, you can even define a nightly version if you want, so even the correct version is ran with rustup. It's fantastic, and I can contribute much easier to projects without worrying about tooling.

languageserver 4y ago
> I don't deal much with C/C++ that is because there is no such thing.
- bee_rider 4y ago
  C/C++ is just 1 in the limit, right?
  - xdavidliu 4y ago
    two things wrong with your statement: - the ++ operator only acts on integer types, not floats or doubles, so there is no limit to speak of here - the expression "C++" has value equal to C before incrementing, hence the expression "C/C++" is just one for positive C, even when C is small
    - bee_rider 4y ago
      So, luckily my mistakes cancel out -- C/C++ = 1 always, so it must also in the limit. Once we figure out how to define limit.
    - Iwan-Zotow 4y ago
      > the ++ operator only acts on integer types no, I believe it works on pointer types and enums as well
      - bee_rider 4y ago
        It is also defined for floats. Using it seems like a bad move though -- for large values it can round back to the input value. Indeed, the following stupid test program works, although it may heat up your laptop slightly. #include <stdio.h> void main() { float C = 1.0f; float Cin = 0.0f; int i=0; while (Cin != C) { i++; Cin = C; C++; } printf("%i %e\n", i, C/C++); } And, it finally lets us confirm what the mathematicians never could. When does the limit happen? 16777216. No further questions.
    - bee_rider 4y ago
      Hey, wait a minute, ++ is defined for floats. It might be a bad idea to use it in many cases (since there are values for which the result is just rounded back to the original value), but it works!
- xdavidliu 4y ago
  the slash symbol is often used to denote multiple entities. For example, "I don't deal with foo / bar" usually means that the commenter doesn't deal with either foo or bar. The commenter is not claiming that foo and bar together constitute a single entity.

languageserver 4y ago

> I don't deal much with C/C++ that is because there is no such thing.

bee_rider 4y ago
C/C++ is just 1 in the limit, right?
- xdavidliu 4y ago
  two things wrong with your statement: - the ++ operator only acts on integer types, not floats or doubles, so there is no limit to speak of here - the expression "C++" has value equal to C before incrementing, hence the expression "C/C++" is just one for positive C, even when C is small
  - bee_rider 4y ago
    So, luckily my mistakes cancel out -- C/C++ = 1 always, so it must also in the limit. Once we figure out how to define limit.
  - Iwan-Zotow 4y ago
    > the ++ operator only acts on integer types no, I believe it works on pointer types and enums as well
    - bee_rider 4y ago
      It is also defined for floats. Using it seems like a bad move though -- for large values it can round back to the input value. Indeed, the following stupid test program works, although it may heat up your laptop slightly. #include <stdio.h> void main() { float C = 1.0f; float Cin = 0.0f; int i=0; while (Cin != C) { i++; Cin = C; C++; } printf("%i %e\n", i, C/C++); } And, it finally lets us confirm what the mathematicians never could. When does the limit happen? 16777216. No further questions.
  - bee_rider 4y ago
    Hey, wait a minute, ++ is defined for floats. It might be a bad idea to use it in many cases (since there are values for which the result is just rounded back to the original value), but it works!
xdavidliu 4y ago
the slash symbol is often used to denote multiple entities. For example, "I don't deal with foo / bar" usually means that the commenter doesn't deal with either foo or bar. The commenter is not claiming that foo and bar together constitute a single entity.