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> Yeah, and it's trivial in C++, what's your point? I see lots of assertions of that fact, but no examples. Care to link some actual code? > Why the hell woul

> Yeah, and it's trivial in C++, what's your point? I see lots of assertions of that fact, but no examples. Care to link some actual code? > Why the hell would you want to change the direction of a pin anyway? Just declare it as Pin<Input> or Pin<Output>. i2c uses the same pin for sending and receiving and changes the mode of the pin for the purposes of communication with peripherals.

IshKebab 3y ago
Ignoring the fact that nobody is bit bashing I2C (there are other protocols that you would bit bash), you don't normally change the mode in that case either. You set it to an output with drive-low and float-high. It's essentially an InputOutput mode.
- sophacles 3y ago
  I'm sure you mean "nobody" not nobody, otherwise this doesn't exist: https://github.com/bitbank2/BitBang_I2C https://github.com/bitbank2/BitBang_I2C (and plenty of others, this was just the first github link google gave me). There's quite a few articles suggesting to bit bang i2c on tiny MCUs also for various memory/code size/etc reasons. I suspect trivial to do in C++ really should be "trivial" too.
  - IshKebab 3y ago
    Yes in that context it means approximately nobody. In English it's normal to not speak 100% literally. Otherwise words like "nobody" and "everyone" become useless because there's always that one guy. The degree of literalness depends on the context, e.g. "everyone hates brutalist architecture" means maybe 90% of people hate it, while "everybody goes through puberty" means more like 99.99%. I'm guessing you're not a native English speaker - surely it is the same in other languages?
    - sophacles 3y ago
      I'm being pedantic since you consistently refuse to back your claims with evidence. I assume you are going to keep deflecting and picking on nonsense instead of just providing the trivial code example. I'm sure you can do it, it's trivial right?
      - IshKebab 3y ago
        Do what? This typestate pattern in C++? Of course you can. It's almost identical to the Rust code. The only downside is that C++'s moves leave the moved-from object accessible but invalid while the Rust compiler will not let you access them. Otherwise it's the same. If you just Google "C++ typestate" the first result is an HTTP library that uses this technique (plus a load of complex template stuff, but that isn't required). But I guess you've seen that already so maybe that's not what you're asking for?
        morelisp 3y ago
        > The only downside is that C++'s moves leave the moved-from object accessible but invalid while the Rust compiler will not let you access them. "Except for the whole damn point, it's the same." Oh, ok.
        IshKebab 3y ago
        That's not the whole damn point. It's arguable half of it - the creation of new states is identical. It's just the destruction that is relegated from compile time to run time.
        morelisp 3y ago
        Yes in that context "whole damn point" means the key, salient aspect of the discussion being only allowing use of the thing once. In English it's normal to not speak 100% literally. I'm guessing you're not a native English speaker - surely it is the same in other languages?
codr7 3y ago
Ok, but then the pin is always both Input/Ouput, so still static; and the example is still as contrived as ever.
- sophacles 3y ago
  Well no, its output when writing and input when reading. I2C isn't full-duplex.

IshKebab 3y ago

Ignoring the fact that nobody is bit bashing I2C (there are other protocols that you would bit bash), you don't normally change the mode in that case either. You set it to an output with drive-low and float-high. It's essentially an InputOutput mode.

sophacles 3y ago
I'm sure you mean "nobody" not nobody, otherwise this doesn't exist: https://github.com/bitbank2/BitBang_I2C https://github.com/bitbank2/BitBang_I2C (and plenty of others, this was just the first github link google gave me). There's quite a few articles suggesting to bit bang i2c on tiny MCUs also for various memory/code size/etc reasons. I suspect trivial to do in C++ really should be "trivial" too.
- IshKebab 3y ago
  Yes in that context it means approximately nobody. In English it's normal to not speak 100% literally. Otherwise words like "nobody" and "everyone" become useless because there's always that one guy. The degree of literalness depends on the context, e.g. "everyone hates brutalist architecture" means maybe 90% of people hate it, while "everybody goes through puberty" means more like 99.99%. I'm guessing you're not a native English speaker - surely it is the same in other languages?
  - sophacles 3y ago
    I'm being pedantic since you consistently refuse to back your claims with evidence. I assume you are going to keep deflecting and picking on nonsense instead of just providing the trivial code example. I'm sure you can do it, it's trivial right?
    - IshKebab 3y ago
      Do what? This typestate pattern in C++? Of course you can. It's almost identical to the Rust code. The only downside is that C++'s moves leave the moved-from object accessible but invalid while the Rust compiler will not let you access them. Otherwise it's the same. If you just Google "C++ typestate" the first result is an HTTP library that uses this technique (plus a load of complex template stuff, but that isn't required). But I guess you've seen that already so maybe that's not what you're asking for?
      - morelisp 3y ago
        > The only downside is that C++'s moves leave the moved-from object accessible but invalid while the Rust compiler will not let you access them. "Except for the whole damn point, it's the same." Oh, ok.
        IshKebab 3y ago
        That's not the whole damn point. It's arguable half of it - the creation of new states is identical. It's just the destruction that is relegated from compile time to run time.
        morelisp 3y ago
        Yes in that context "whole damn point" means the key, salient aspect of the discussion being only allowing use of the thing once. In English it's normal to not speak 100% literally. I'm guessing you're not a native English speaker - surely it is the same in other languages?

codr7 3y ago

Ok, but then the pin is always both Input/Ouput, so still static; and the example is still as contrived as ever.

sophacles 3y ago
Well no, its output when writing and input when reading. I2C isn't full-duplex.