What is the reason behind this code's ability to detect vowels as well as other

Question

What is the reason behind this code's ability to detect vowels as well as other

I'm currently working on debugging a specific portion of code for my class. I'm having trouble understanding why this JavaScript code is counting all letters instead of just vowels.

var text, i, sLength, myChar;
var count;
var text = prompt("Please type a phrase"); //changed to "phrase" and fixed ' to "
count = 0;
for (i = 1; i <= text.length; i+= 1){
    myChar = text[i];
    if (myChar == 'a' || myChar == 'o' || myChar == 'e' || myChar == 'u'){ //modified the conditionals for checking vowels
        count += 1;
        console.log('Vowel:', myChar);   
    }
    console.log(myChar, count);
}
alert(count); //moved count into ()

javascript debugging

Answer 1

Answer №1

The comparison line is not executing the check correctly. It should read as follows:

if (myChar == 'a' || myChar == 'o' || myChar == 'e' || myChar == 'u')

Answer 2

The comparison line is not executing the check correctly. It should read as follows:

if (myChar == 'a' || myChar == 'o' || myChar == 'e' || myChar == 'u')

Answer 3

Answer №2

Here is the corrected JavaScript code:

var text, i, sLength, myChar;
var count;
var text = prompt("Type a phrase please");
count = 0;
for (i = 1; i <= text.length; i+= 1){
    myChar = text[i];
    if (myChar == 'a' || myChar == 'o' || myChar =='e' || myChar == 'u'){ 
        count += 1;
        console.log('Vowel:', myChar);   
    }
    console.log(myChar, count);
}
alert(count)

The issue with your original code is that it incorrectly evaluates whether 'o' is true in the expression || 'o'. Since a string literal like 'o' is not a falsy value, the condition stays true for every iteration of the loop.

Answer 4

Here is the corrected JavaScript code:

var text, i, sLength, myChar;
var count;
var text = prompt("Type a phrase please");
count = 0;
for (i = 1; i <= text.length; i+= 1){
    myChar = text[i];
    if (myChar == 'a' || myChar == 'o' || myChar =='e' || myChar == 'u'){ 
        count += 1;
        console.log('Vowel:', myChar);   
    }
    console.log(myChar, count);
}
alert(count)

The issue with your original code is that it incorrectly evaluates whether 'o' is true in the expression || 'o'. Since a string literal like 'o' is not a falsy value, the condition stays true for every iteration of the loop.

Answer 5

Answer №3

Quick and Easy Vowel Count

function countVowels (string) {
    return string.match(/a|e|i|o|u/g).length;
}

Try it out like this:

alert(countVowels(prompt("Enter a phrase")));

Your if statement is not valid. For a simple check on multiple options, consider using the following method:

if (['a', 'e', 'i', 'o', 'u'].indexOf(myChar) > -1) {
}

Another Handy Function!

function hasOptions (c) {
    return arguments.splice(1).indexOf(c);
}

Then

if (hasOptions(myChar, 'a', 'e', 'i', 'o', 'u'))

Answer 6

Quick and Easy Vowel Count

function countVowels (string) {
    return string.match(/a|e|i|o|u/g).length;
}

Try it out like this:

alert(countVowels(prompt("Enter a phrase")));

Your if statement is not valid. For a simple check on multiple options, consider using the following method:

if (['a', 'e', 'i', 'o', 'u'].indexOf(myChar) > -1) {
}

Another Handy Function!

function hasOptions (c) {
    return arguments.splice(1).indexOf(c);
}

Then

if (hasOptions(myChar, 'a', 'e', 'i', 'o', 'u'))

What is the reason behind this code's ability to detect vowels as well as other

Answer №1

Answer №2

Answer №3

Quick and Easy Vowel Count

Another Handy Function!

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