6 ms·

I'm still puzzled to why would a compiler allow assignment in an explicit conditional (outside of loop syntax). It's like a baked-in blindspot that most people

• arkadiytehgraet 8y ago
  Funny you should mention Kotlin; while I like the language a lot, I believe this particular feature would be of immense help in the following scenario: Imagine you have a sealed class Foo, with `class Bar(x: String) : Foo()` and `Baz() : Foo()`. Now, imagine you have a method, returning an object of type Foo: `fun foo(): Foo` And you want to pattern match on the result of this method: when(foo()) { is Bar -> ... is Baz -> ... } Now, the problem is: how do you access String field x in the first branch? The only way to do it now is to extract the methos call into redundant local variable, and then pattern match it instead of `foo()` directly as I did above. Now imagine Kotlin had that feature; then we could just do the following: when(foo = foo()) { is Bar -> foo.x is Baz -> ... }
  • bpicolo 8y ago
    One option is match statements. Great way to make this sort of inline assignment unnecessary
    • arkadiytehgraet 8y ago
      Could you please elaborate more on what you mean by match statements? Is it an already existing feature of Kotlin?
      • bpicolo 8y ago
        Ah, I guess `when` is literally Kotlin's equivalent of match. What I'm thinking of here is Rust's match statements, which do give you the ability to make use of those intermediary values by making use of Rust's enum type. https://doc.rust-lang.org/book/second-edition/ch06-02-match.html#matching-with-optiont https://doc.rust-lang.org/book/second-edition/ch06-02-match....
        • arkadiytehgraet 8y ago
          I see; I agree that proper pattern matching would indeed solve that as well, as e.g. Scala does.
  • bjz_ 8y ago
    In Rust and Haskell you can do: match foo() { Bar @ foo => foo.x, Baz => ..., }
• kelnos 8y ago
  Because assignment is an expression. It has nothing specifically to do with the position being a conditional. Conditionals need an expression, and assignment fits the bill, so it works.
  • KSteffensen 8y ago
    Why is assignment an expression and not a statement? Why should an assignment evaluate to a value?
  • TooBrokeToBeg 8y ago
    > Because assignment is an expression. It has nothing specifically to do with the position being a conditional. No, but it has a lot to do with the way humans write code. This is the source of the bug, not the logic. eg Why bother having whitespace that the compiler can't use (typically)? Human readability. The concession to not take into account human failing, is pathological.
  • hinkley 8y ago
    The other partial fix is to stop allowing truthy conditionals. Only allow Boolean values and the only time you it wrong is when you’re assigning to booleans.
• shakna 8y ago
  It's handy in C for guarding parts of the language, as you often have to check return values anyway. if(malloc(10000 * sizeof foo) == NULL) { // Error processing } // No need for an else branch here.
  • the-dude 8y ago
    Missing assingment.
    • shakna 8y ago
      int success = NULL; if(success = malloc(...)) { } if(success = malloc(...)) { } if(success = malloc(...)) { }
      • slavik81 8y ago
        That will result in undefined behaviour if the value returned by malloc is greater than INT_MAX. If you really need an integer, use intptr_t. But, generally it makes more sense to just use a pointer. char* p = NULL; if (p = malloc(...)) { ... }
• wool_gather 8y ago
  Well, if you have the concept of Maybe/Optional, it's quite handy to have a conditional binding as control flow (as in Swift): guard let value = myOptional else { // Handle absence } // Use value Or if let value = myOptional { // Use value }
• emmelaich 8y ago
  Read Tim Peter's essay on the end of the PEP 572. It's very good.