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please help

•

Join Date: Aug 2007

Posts: 4

Reputation: angelfox is an unknown quantity at this point

Solved Threads: 0

angelfox

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Newbie Poster

please help

Aug 29th, 2007

i know this is 2 easy for you guys, but please help..... Its been only 2 wks since i enrolled this class and they gave me this assignment.... please check if its correct. thank you so much

 Help with Code Tags 
 cpp Syntax (Toggle Plain Text) 
 #include<iostream>
using namespace std;
 
char vowels[]={a,e,i,o,u};
char n;
 
char main()
{
cout<<“Please Enter a Character:\n”;
cin>>n;
{
if (n=vowels)
       cout<<“n is a VOWEL\n”;
else
       cout<<“n is CONSONANT\n”;
        }
return 0;
}
 #include<iostream>
using namespace std;

char vowels[]={a,e,i,o,u};
char n;

char main()
{
cout<<“Please Enter a Character:\n”;
cin>>n;
{
if (n=vowels)
       cout<<“n is a VOWEL\n”;
else
       cout<<“n is CONSONANT\n”;
        }
return 0;
}

Last edited by WolfPack; Aug 29th, 2007 at 12:13 pm. Reason: ADDED [CODE=CPP][/CODE] TAGS

•

Join Date: Dec 2006

Posts: 44

Reputation: Boldgamer is an unknown quantity at this point

Solved Threads: 2

Boldgamer

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Light Poster

Re: please help

Aug 29th, 2007

I am a begginer also but I believe you need main to be declared as

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int main()
int main()

The line

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
 char vowels[]={a,e,i,o,u};
 char vowels[]={a,e,i,o,u};

isn't needed.
I would put

Help with Code Tags

C++ Syntax (Toggle Plain Text)

 char n;

inside int main().
For your branching statement you would declare each of the vowels in the condition statement i.e.

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
if (n == 'A' || n== 'a' || n == 'E' || ect...)
if (n == 'A' || n== 'a' || n == 'E' || ect...)

It would probably be a lot easier to use a switch statement though. i.e.

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
 
    switch (n)
    {
    case 'A':
    case 'a':
    case 'E':
    case 'e':
    case 'I':
    case 'i':
    case 'O':
    case 'o':
    case 'U':
    case 'u':
          {
              cout << "This letter is a Vowel." << endl;
              break;
          }
    default:
        {
            cout << "This letter is a consonant." << endl;
        }
    }
 
    switch (n)
    {
    case 'A':
    case 'a':
    case 'E':
    case 'e':
    case 'I':
    case 'i':
    case 'O':
    case 'o':
    case 'U':
    case 'u':
          {
              cout << "This letter is a Vowel." << endl;
              break;
          }
    default:
        {
            cout << "This letter is a consonant." << endl;
        }
    }

Only problem with this is that if someone enters something that is not an alphabet it will be considered a consonant.

Last edited by Boldgamer; Aug 29th, 2007 at 12:33 pm.

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Join Date: Jul 2005

Posts: 1,671

Reputation: Lerner is a name known to all

Solved Threads: 261

Lerner

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Posting Virtuoso

Re: please help

Aug 29th, 2007

Welcome to Daniweb. I see that Wolfpack helped formating your code since it was your first post. In the future please use code tags when posting information, like code, where you want to preserve indentation.

As another alternative you could use the array and a loop to read through array to check if the given char is a vowel or not. Of course this assumes you know about accessing elements of an array and control loops, etc.

Note how Boldgamer refers to single chars using single quotes before and after the char, and how Bolgamer used the == operator rather than the = operator for equals.

Trying to avoid global variables is considered good form in standard C++.

Last edited by Lerner; Aug 29th, 2007 at 12:41 pm.