In VS I am told this function “does not return a value in all control paths.” A bot told me specifically the issue is with this line: else if (letter + key <= 90). It said that if the outcome results in letter + key equally exactly 90 then a value is not returned, but I thought that was covered where ‘<=’ means ‘less than or equals.’

char rotate(char letter, int key)
{
    if (isalpha(letter) == true)
    {
        if (letter + key > 90)
        {
            int overage = letter + key - 90;
            letter = 64 + overage;

            while (letter > 90)
            {
                overage = letter - 90;
                letter += overage;
            }

            return letter;
        }

        else if (letter + key &lt;= 90)
        {
            letter += key;
            return letter;
        }
    }

    else if (isalpha(letter) == false)
    {
        return letter;
    }
  • my_hat_stinks
    link
    711 months ago

    That’s because they’re not necessarily mutually exclusive. The function is being called twice so there’s no way to guarantee the result will be the same both times without knowing what it does under the hood.

    Consider a case where isalpha performs a coin flip, 50% chance each call to return true. The first call returns false so the first condition fails, then the second call returns true so the second condition fails; in 25% of cases neither code block executes.

    You could store the result of the first call in a local variable and reuse it if you really wanted to, but the smart solution is to either use if/else properly or switch to early returns instead.

    • @[email protected]
      link
      fedilink
      211 months ago

      Right, the compiler isn’t smart enough to recognize that isalpha() is pure and deterministic.

    • @mrkite
      link
      111 months ago

      Expect isalpha is part of the standard library not an arbitrary function, a compile should be able to optimize standard calls.

      • my_hat_stinks
        link
        111 months ago

        Compiler optimisations don’t apply when you’re breaking the rules of the language. It won’t compile.