Programming

17366 readers

495 users here now

Welcome to the main community in programming.dev! Feel free to post anything relating to programming here!

Cross posting is strongly encouraged in the instance. If you feel your post or another person's post makes sense in another community cross post into it.

Hope you enjoy the instance!

Rules

Follow the programming.dev instance rules
Keep content related to programming in some way
If you're posting long videos try to add in some form of tldr for those who don't want to watch videos

Wormhole

Follow the wormhole through a path of communities !webdev@programming.dev

founded 1 year ago

MODERATORS

snowe@programming.dev

Ategon@programming.dev

MaungaHikoi@lemmy.nz

Why does a control path not execute in this function? (lemm.ee)

submitted 11 months ago by milon@lemm.ee to c/programming@programming.dev

12 comments fedilink hide all child comments

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;
    }

you are viewing a single comment's thread
view the rest of the comments

[–] JakenVeina@lemm.ee 3 points 11 months ago (1 children)

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.

[–] my_hat_stinks@programming.dev 5 points 11 months ago (2 children)

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.

[–] JakenVeina@lemm.ee 2 points 11 months ago

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

[–] mrkite@programming.dev 1 points 11 months ago (1 children)

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

[–] my_hat_stinks@programming.dev 1 points 11 months ago

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