4 ms·

I agree, although I suppose different operators could still have different associativity in a language without precedence. I took that phrasing from the “Puffs

• masklinn 9y ago
  > I suppose different operators could still have different associativity in a language without precedence That… is true, and I have a hard time computing how it would even work.
  • klodolph 9y ago
    Think exponentiation, where 2^3^4 might be 2^(3^4), but where subtraction 2-3-4 is (2-3)-4.
    • mnarayan01 9y ago
      Consider: 2-3^4
      • irishsultan 9y ago
        Since there is no precedence in this (hypothetical) language that would be an invalid expression.
  • evincarofautumn 9y ago
    One way to do it would be to have a default associativity between different operators, but chains of the same operator can override it, e.g. suppose “⧺” means concatenation (rightward), “×” means repetition (leftward), and the default is leftward: [1, 2] ⧺ [3, 4] ⧺ [5, 6] × 2 × 3 (([1, 2] ⧺ ([3, 4] ⧺ [5, 6])) × 2) × 3 [ [ [1…6], [1…6] ], [ [1…6], [1…6] ], [ [1…6], [1…6] ] ] I dunno how good an idea that would be in practice, though.