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 <= 90)
{
letter += key;
return letter;
}
}
else if (isalpha(letter) == false)
{
return letter;
}
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.
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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
isalphaperforms 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.