C++ show vowel and consonant and count it

Question

When user input number from 1 - 26 (which mean a to z), how to show the the letter and count how many vowel and consonant inside it.

Example: U

Answer 1

Your if statement does not do what you expect it to. The correct syntax for this is

if (let=='a' || let=='e' || let=='i' || let=='o' || let=='u')

The reason the current versions is incorrect is because it is equivalent to

if ((let=='a') or ('e') or ('i') or ('o') or ('u'))

So 'e', 'i', etc are being evaluated for truthiness and ord to the first condition. Only an empty string in this case will evaluate to false, all of these characters will be true. So your statement evaluates to

if ((let=='a') or true or true or true or true)

Which will always be true.