C++ Problem Solving Scenarios - Urgent Help Needed Please

[StacyG]

1. Write a program that asks the user to input an email address. Check whether the email address is valid or not. The conditions to be checked are:
• Email address should not contain more than 10 characters.
• Email address should not contain spaces.
• Email address should contain @ in it.
• The last three characters should either be “com” or “edu” or “org”.



#include <iostream>
#include <string>
using namespace std;

int main()
{
string emailaddress;

cout <<"Please enter an email address :";
getline(cin, emailaddress);

int elenght = emailaddress.length();
if (elenght > 10)
{
cout <<"Your email must NOT contain more than 10 characters, please try again"<<endl;
return 0;
}

int sp = emailaddress.find(" ");
if (sp > 0)
{
cout << "Your email must NOT contain any spaces, please try again" << endl;
return 0;
}

int at = emailaddress.find("@");
if (at == string::npos)
{
cout << "Your email MUST contain an '@' sign, please try again" << endl;
return 0;
}


int end1 = emailaddress.rfind("com", "edu", "org" );
if (end1 == string::npos)
{
cout << "Please enter an email domain from the following list - com, edu, org" << endl;
return 0;
}

else
cout<< emailaddress << " Email Address not valid";

return 0;
}

[Ganado]

Show us what you've tried. If you show some effort, we can point out where mistakes are. We can't know what's wrong if you don't show any code.

http://www.cplusplus.com/doc/tutorial/

[StacyG]

Sorry about that. The part below is for the email address not containing more than 10 characters:

#include <iostream>
#include <string>
using namespace std;

int main()

{

std:: string email_address [10];
std:: string email = "Please enter your Email ";
std:: string email_address = email + "Address:";
std:: cout << email_address <<std::endl;


}

[Ganado]

A string by itself is already a series of characters, so the [10] in your code is actually making an array of 10 strings, which is not what you want.

Just declare it as:
std::string email_address;

And remove your third line. It doesn't make sense, you're re-declaring email_address.

Use std::cin to get input from the user.
See: http://www.cplusplus.com/doc/tutorial/basic_io/

To check the length of a string, use email_address.length()
You're going to want if statements to verify the rules of what you're checking for,
see: http://www.cplusplus.com/doc/tutorial/control/

To check if a string contains a particular character, you can do

std::string my_str = "blah@bleh";
if (my_str.find("@") != std::string::npos)
{
    // at sign found
}
else
{
    // at sign not found
}

[StacyG]

Thank you so much, will follow your suggestions and see if I can get this running. Thank you again!

[jonnin]

1. Write a program that asks the user to input an email address. Check whether the email address is valid or not.

ironically that is a very difficult real world challenge that manager/nontechnical people can't understand just how hard it actually is. The rules you have are not correct in real life by a long shot. Its a good real world problem to think about, even with the simplified rule set.

there is a substring function that can take the last 3 letters, then you can just check equality.

@@@.edu appears to be legal with your rule set. Hehe.
so is derp@edu
is upper case EDU etc ok? may want to watch for case on those, or ask your prof.

[thmm]

@jonnin,
I guess the task is more about string processing than real email validation.

[dhayden]

A few hints:

• Email address should not contain more than 10 characters.

The string class has a size() method that returns the number of characters in the string.

• Email address should not contain spaces.

The string class has a find() method that will find the position of a character in a string. It returns string::npos if the character isn't found.

• Email address should contain @ in it.

Use the find() method again.

• The last three characters should either be “com” or “edu” or “org”.

The string class has a substr() class that returns part of a string. Use it to extract the last 3 characters. You'll need to do a little math to figure out how to get the last 3.
http://www.cplusplus.com/reference/string/string/substr/

[StacyG]

Thanks everyone, here is what I did so far, it's not complete but it's a start:

#include <iostream>
#include <string>
using namespace std;

int main()
{
string emailaddress;

cout <<"Please enter an email address :";
getline(cin, emailaddress);

int elenght = emailaddress.length();
if (elenght > 10)
{
cout <<"Your email must NOT contain more than 10 characters, please try again"<<endl;
return 0;
}

int sp = emailaddress.find(" ");
if (sp > 0)
{
cout << "Your email must NOT contain any spaces, please try again" << endl;
return 0;
}

int at = emailaddress.find("@");
if (at == string::npos)
{
cout << "Your email MUST contain an '@' sign, please try again" << endl;
return 0;
}


int end1 = emailaddress.rfind("com", "edu", "org" );
if (end1 == string::npos)
{
cout << "Please enter an email domain from the following list - com, edu, org" << endl;
return 0;
}

else
cout<< emailaddress << " Email Address not valid";

return 0;
}

[jonnin]

I guess the task is more about string processing than real email validation.

Of course it is. I was just commenting that its a deceptively difficult problem in reality.

OP, that is a good start. you can combine statements to save pointless clutter and cpu work, though.

int elenght = emailaddress.length();
if (elenght > 10)
{

consider instead

if (emailaddress.length() > 10)
{

[Ganado]

In my opinion, that shouldn't be what a beginner needs to worry about; in fact both versions would probably be compiled to the same thing.

StacyG, what's more flagrant is int end1 = emailaddress.rfind("com", "edu", "org" );
Did you read documentation for rfind? I don't even think that would compile.

C++20 is (finally) introducing an ends_with function: https://en.cppreference.com/w/cpp/string/basic_string/ends_with

However, I assume you can't use that.
You need to use something equivalent to ends_with("edu") || ends_with("com") etc.

The easiest way to do this, from what I can tell, is to take the substring of the last three characters.
Look at this example:

// Example program
#include <iostream>
#include <string>
using namespace std;

string get_last_n_chars(const string& str, int n)
{
    if ((int)str.length() < n)
        return "";
        
    return str.substr(str.length() - n);
}

int main()
{
    string str = "another happy landing";
    
    cout << get_last_n_chars(str, 3) << '\n';

    if (get_last_n_chars(str, 7) == "landing")
    {
        cout << "Ends with landing\n";
    }
}

ing
Ends with landing

[jonnin]

In my opinion, that shouldn't be what a beginner needs to worry about; in fact both versions would probably be compiled to the same thing.

the visual clutter and extra typing are worth cutting out even as a beginner. Its minor and not the focus, agreed, but the less clutter in code, the easier it is to read and debug, at least to me.

[dhayden]

StacyG, nice code! It's refreshing to see a beginner who catches on so quickly.

Please use code tags when posting. Edit your post, highlight the code and click the "<>" button to the right of the edit window.

int sp = emailaddress.find(" ");
if (sp > 0)
{
cout << "Your email must NOT contain any spaces, please try again" << endl;
return 0;
}

This has a couple of problems. First, what if the string starts with a space? Remember that in C++, indexes start with 0, not 1.

Also, find() will return the magic number string::npos if the value isn't found. What if that number is positive? The only reliable way to test if something is found is to compare it to string::npos, so the if should be :
if (sp != string::npos)

Also, because find returns a number of type size_t, you should declare sp as a size_t:
size_t sp = emailaddress.find(" ");

After this, the only change needed is the one discussed above to compare against the last 3 characters.

[StacyG]

Thank you for your suggestions. Will do some more reading to better understand.