Thread: Strings and stuff

  1. #1
    casanova0o7
    Guest

    Unhappy Strings and stuff

    Hi all,
    I'm having trouble diagnosing a while loop I created to analyze a string's number of vowels and consonants. Here is my code:


    while (string[i] != '\n')
    {
    if (string[i] == 'a' )
    vowels++;
    else
    consonants++;

    i++;

    }//end of analyzing loop

    Some help would be greatly appreciated.

  2. #2
    &TH of undefined behavior Fordy's Avatar
    Join Date
    Aug 2001
    Posts
    5,793
    For a start, the condition tests for '\n' not '\0' or NULL.....

    What other problems have you got?

  3. #3
    casanova0o7
    Guest

    Counting vowels

    I think I need one more tweak to make the whole engine work. The change from '/n' to '/0' worked, but the loop doesn't separate the vowels from the consonants. Instead, it counts all the characters as vowels.

    Here is the code I used:

    char string[256];
    int vowels=0, consonants=0;

    cout << "Input string: ";
    cin.getline( string, 256, '\n');

    cout << "You typed: " << string << endl;

    int i=0;

    while (string[i] != '\0')
    {
    if (string[i] == 'a' || 'e' || 'i' || 'o' || 'u' )
    vowels++;
    else
    consonants++;
    i++;

    }//end of analyzing loop
    char string[256];
    int vowels=0, consonants=0;

    cout << "Input string: ";
    cin.getline( string, 256, '\n');

    cout << "You typed: " << string << endl;

    int i=0;

    while (string[i] != '\0')
    {
    if (string[i] == 'a' || 'e' || 'i' || 'o' || 'u' )
    vowels++;
    else
    consonants++;

    i++;

    }//end of analyzing loop

    cout << "Number of vowels in your string: " << vowels << endl;
    cout << "Number of non-vowels in your string:" << consonants << endl;

  4. #4
    ¡Amo fútbol!
    Join Date
    Dec 2001
    Posts
    2,138
    Code:
    if (string[i] == 'a' || 'e' || 'i' || 'o' || 'u' )
    not legal!!!

    Have to include seperate equals statements. Use a newline for each vowel comparison to make ur code look neat.

    if (string[i]=='a' ||
    string[i]=='b' ||

    and so on...

  5. #5
    casanova0o7
    Guest
    Good call! I had a similar line work in a different application (with int) so I never suspected that illegality. Thanks!

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