What is wrong with my code that removes vowels from strings? - c

Question

I'm new to programming and I've decided to do some simple programming exercises.
The challenge is to remove all vowels from a user inputted string.
I've already wrote the code and I don't know why it isn't working.
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <ctype.h>
bool isVowel(char ch)
{
char charToBeTested = tolower(ch);
if(charToBeTested == 'a' || 'e' || 'i' || 'o' || 'u')
return true;
else
return false;
}
int main()
{
int i;
char formattedString[80];
printf("Enter a string: ");
scanf("%s", &formattedString);
for(i = 0; i < strlen(formattedString); i++)
{
if(isVowel(formattedString[i]) == true)
{
formattedString[i] = ' ';
}
}
printf("%s", formattedString);
return 0;
}
All it is supposed to do is check every character in the string and see if it's a vowel. If it's a vowel, replace the current character with a space. I will write the function to remove the spaces later.
All help is appreciated, sorry for being such a noob!

Answer 1

This code doesn't do what you think it does:
if (charToBeTested == 'a' || 'e' || 'i' || 'o' || 'u') {
...
}
C interprets this as
if ((charToBeTested == 'a') || 'e' || 'i' || 'o' || 'u') {
...
}
Here, the || operator is being applied directly to the character literals 'e', 'i', etc. Since C treats any nonzero value as "true," this statement always evaluates to true.
To fix this, try rewriting it like this:
if (charToBeTested == 'a' ||
charToBeTested == 'e' ||
charToBeTested == 'i' ||
charToBeTested == 'o' ||
charToBeTested == 'u') {
...
}
Alternatively, use a switch statement:
switch (charToBeTested) {
case 'a': case 'e': case 'i': case 'o': case 'u':
return true;
default:
return false;
}
Also, you might want to use tolower so that the testing is done case-insensitively.
Hope this helps!

Answer 2

This is incorrect:
if(charToBeTested == 'a' || 'e' || 'i' || 'o' || 'u')
correct is:
if(charToBeTested == 'a' || charToBeTested == 'e' || charToBeTested == 'i' || charToBeTested == 'o' || charToBeTested == 'u')
Or, you can create static table, like:
char vowels[0x100] = {
['a'] = 1,
['e'] = 1,
['i'] = 1,
['o'] = 1,
['u'] = 1,
};
And test by:
if(vowels[(unsigned char)charToBeTested])

Answer 3

You'll need to rewrite your isVowel function to something like:
bool isVowel(char ch)
{
char charToBeTested = tolower(ch);
if(charToBeTested == 'a') {
return true;
} else if(charToBeTested == 'e') {
return true;
} else if(charToBeTested == 'i') {
return true;
} else if(charToBeTested == 'o') {
return true;
} else if(charToBeTested == 'u') {
return true;
} else {
return false;
}
}
Alternatively, you can use a switch statement to do the same thing, like this:
bool isVowel(char ch)
{
char charToBeTested = tolower(ch);
switch(charToBeTested) {
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
return true;
default:
return false;
}
}

Answer 4

You need to test each alternative separately:
char c = tolower(ch);
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
Your original code is:
if (charToBeTested == 'a' || 'e' || 'i' || 'o' || 'u')
The compiler evaluates that as:
if ((charToBeTested == 'a') || true)
because 'e' is not zero and any expression which is not zero is true.
If it optimizes really thoroughly, can deduce that the whole expression will always be true, regardless of the value of charToBeTested, and hence it can reduce the entire function to return true (without ever needing to call tolower(). If it was a static function, it could even eliminate the function call altogether. Whether any compiler would actually be that aggressive is open to debate.

Answer 5

Just to add to everyones answer; as they say, you need to fix your testing criteria.
Also, don't forget to check for the Uppercase version too:
if (charToBeTested == 'a' ||
charToBeTested == 'e' ||
charToBeTested == 'i' ||
charToBeTested == 'o' ||
charToBeTested == 'u' ||
charToBeTested == 'A' ||
charToBeTested == 'E' ||
charToBeTested == 'I' ||
charToBeTested == 'O' ||
charToBeTested == 'U') {
...
}