3 ms·

I'm a bit confused - isn't this how all Static Single Assignment representations in compilers work? And those are used in things like LLVM IR to represent C and

I'm a bit confused - isn't this how all Static Single Assignment representations in compilers work? And those are used in things like LLVM IR to represent C and C++ code. Is C++ immutable now?

kqr 2y ago
The difference at the high level is that assigning a variable creates a new scope. E.g. in C I would expect to be able to int i = 0; while (i < 5) { i = i+1; printf("i: %d\n", i); } whereas in Haskell I could hypothetically something like let i = 0 in whileM (pure (i < 5)) $ let i = i + 1 in printf "i: %d\n" i but the inner assignment would not have any effect on the variable referenced by the condition in the while loop – it would only affect what's inside the block it opens. (And as GP points out, i=i+1 is an infinite loop in Haskell. But even if it was used to build a lazy structure, it would just keep running the same iteration over and over because when the block is entered, i still has the value that was set outside.)
- eru 2y ago
  Btw, Haskell also supports mutable re-assignment of variables. But it's not something that's built into the language, you get mutable variables via a library. Just like you can get loops in Haskell via a library.
  - kqr 2y ago
    Oh, yeah, it's right there in the standard library. But one has to be a little more explicit about the fact that one is accessing truly mutable variables. For example, given the helper utilities checkIORef r p = fmap p (readIORef r) usingIORef r a = readIORef r >>= a the example can be written as main = do i <- newIORef (0 :: Int) whileM_ (checkIORef i (< 5)) $ do modifyIORef i (+1) usingIORef i (printf "i: %d\n") That said, even if one actually needs to mix mutable variables with I/O actions like printing, I'm not sure I would recommend using IORefs for it. But opening the can of MonadIO m => StateT Int m () is for another day.
    - eru 2y ago
      You can also use the State monad, instead of IORef.
      - kqr 2y ago
        Not while printfing. That is what takes a MonadIO m => StateT s m () transformer, or at least a stack with a Writer in it.
        eru 2y ago
        Yes, that's true.
deleted 2y ago
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I'm a bit confused - isn't this how all Static Single Assignment representations in compilers work? And those are used in things like LLVM IR to represent C and C++ code. Is C++ immutable now?

kqr 2y ago

The difference at the high level is that assigning a variable creates a new scope. E.g. in C I would expect to be able to int i = 0; while (i < 5) { i = i+1; printf("i: %d\n", i); } whereas in Haskell I could hypothetically something like let i = 0 in whileM (pure (i < 5)) $ let i = i + 1 in printf "i: %d\n" i but the inner assignment would not have any effect on the variable referenced by the condition in the while loop – it would only affect what's inside the block it opens. (And as GP points out, i=i+1 is an infinite loop in Haskell. But even if it was used to build a lazy structure, it would just keep running the same iteration over and over because when the block is entered, i still has the value that was set outside.)

eru 2y ago
Btw, Haskell also supports mutable re-assignment of variables. But it's not something that's built into the language, you get mutable variables via a library. Just like you can get loops in Haskell via a library.
- kqr 2y ago
  Oh, yeah, it's right there in the standard library. But one has to be a little more explicit about the fact that one is accessing truly mutable variables. For example, given the helper utilities checkIORef r p = fmap p (readIORef r) usingIORef r a = readIORef r >>= a the example can be written as main = do i <- newIORef (0 :: Int) whileM_ (checkIORef i (< 5)) $ do modifyIORef i (+1) usingIORef i (printf "i: %d\n") That said, even if one actually needs to mix mutable variables with I/O actions like printing, I'm not sure I would recommend using IORefs for it. But opening the can of MonadIO m => StateT Int m () is for another day.
  - eru 2y ago
    You can also use the State monad, instead of IORef.
    - kqr 2y ago
      Not while printfing. That is what takes a MonadIO m => StateT s m () transformer, or at least a stack with a Writer in it.
      - eru 2y ago
        Yes, that's true.

deleted 2y ago

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