1y ago

Why does a control path not execute in this function?

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;
    }

12 comments

you are missing a returning else block.
- Ah I see. I had a bad habit of using else if statements instead of else statements because I thought else if could be better in seeing the condition it's testing for so it was clearer. I get the logic is actually different now.
  
  I'm a fan of early returns.
  So
  
  if(isalpha(letter) == false) return letter if(letter whatever) { Do thing Return letter } if(letter something else) { Do something else Return letter } throw error("unprocessable letter")
  I find it lets me mentally walk through all the paths more easily.
  And if something gets too complex for this, then I need to break it down into further functions
  
  It's not at all necessary, but I find it makes much easier to read code if you instead only use if statements and just return early when you're in a function. For example, you could check isalpha(letter) == true is true then check letter + key <= 90 do the letter += key; return letter; then since letter + key must be 90 if it didn't already return a value, then you can do the while statement and return letter without needing an if statement at all. Then the isalpha(letter) == false is also unncessary, just return letter at the end.
  Like this:
  
  char rotate(char letter, int key) { if (isalpha(letter) { if (letter + key <= 90) { letter += key; return letter; } do the while loop here } return letter; }
  
  I second the early return suggestions, but the other thing you should be aware of is that isalpha() is being evaluated twice unnecessarily. If its cheap and not call frequently, not a real big problem, but it is a waste of cycles. If you want to document the else, you could try:
  c
  
  ... } else // isAlpha() == false { ...
  Also, if isAlpha was something that could change between evaluation, such as isTuesday, you are at risk of the first call returning false, and then the second call returning true, which would skip both cases.
  
  you are checking the result of a method that returns a boolean (true/false) there is no reason to check for true, else, FILE_NOT_FOUND. you can also forgo equating its return value to something to get a boolean value required for the if statement since its already a boolean value.
  e.g.
  
  if (isalpha(letter)) { // ... } else { // ... }
  
  https://en.wikipedia.org/wiki/Defensive_programming might be an interesting read.

This particular compiler isn't smart enough to recognize that isalpha(letter) == true and isalpha(letter) == false are mutually exclusive conditions. It thinks there's a third scenario that you haven't accounted for.
- 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.
  
  Right, the compiler isn't smart enough to recognize that isalpha() is pure and deterministic.
  
  Expect isalpha is part of the standard library not an arbitrary function, a compile should be able to optimize standard calls.

12 comments