Java: how to use variable in a method

Question

The code inspects the chars in string invemail ("sammy99gmail.com"). notice how the email is missing the '@' sign, this program should tell if it's there or not, if so, print invemail "is a valid email" else invemail "is not valid".

The problem is that i get an error highlighted at 'emailfound' on the second 'if' statement. what am i doing wrong?

code:

String invemail = "sammy99gmail.com";


    for (int i = 0; i < invemail.length(); i++)
    {
        char emailfind = invemail.charAt(i);
        if (emailfind == '@')
        {
            boolean emailfound = true;
        }

        else
        {
            boolean enotfound = false;
        }


    }

    if (emailfound = true)
    {
        System.out.println(invemail + "is a valid email");
    }
    else
    {
        System.out.println(invemail + "is invalid");
    }

You are setting variables which have no scope and you are not checking whether emailfound == true or just if (emailfound) but instead you are assigning emailfound = true and then checking is true is still true. — Peter Lawrey
– Peter Lawrey, Commented Aug 30, 2015 at 15:15

colouredmirrorball · Accepted Answer · 2015-08-30 15:24:05Z

If it can't find an @ sign, the boolean emailfound would not get declared at all! So the compiler gives an error saying it's possible emailfound does not exist at that part of the code. To solve this simply declare the boolean outside the first if statement so that it still exists when the second if statement is reached:

boolean emailfound = false;
for(int i = 0; i < invemail.length(); i++)
//etc

By setting it to false by default, it means you don't need to set it to false when it can't find the @ symbol, so you wouldn't need the else here. Be careful that when you found an @, you should assign true to emailfound without declaring a new one: emailfound = true; instead of doing boolean emailfound = true; as that would result in an error about duplicate local variable.

Another thing I spotted is in the second if statement: if (emailfound = true). Careful here! You used = instead of == which means it will actually assign the value true to emailfound instead of evaluating if it's equal to true! In fact you don't need to compare it at all, the if statement expects a boolean and emailfound is a boolean. It's always better to do if(emailfound) instead of if(emailfound == true) precisely to avoid this mistake.

Peter Lawrey · Accepted Answer · 2015-08-30 15:27:14Z

0

You can write this with a boolean, using a label.

String invemail = "sammy99gmail.com";

found: {
    for (int i = 0; i < invemail.length(); i++) {
        char emailfind = invemail.charAt(i);
        if (emailfind == '@') {
            System.out.println(invemail + "is a valid email");
            break found;
         }
    }
    System.out.println(invemail + "is invalid");
}

You can avoid using a label by defining a method to perform a check.

answered Aug 30, 2015 at 15:27

Peter Lawrey

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1 Comment

Clashsoft Over a year ago

It'd be better practice to put the code in a method rather than using the obscure break statement-block style.

Bruno Caceiro · Accepted Answer · 2015-08-30 16:36:35Z

0

You are declaring emailfoundinside an if statement. By doing this, the variable is one visible inside that statement. Since you are accessing emailfoundin other section of the code, the error you get is because the variable is not globaly declared.

Change to:

String invemail = "sammy99gmail.com";
   boolean emailfound = false;
   boolean enotfound = false;

    for (int i = 0; i < invemail.length(); i++)
    {
        char emailfind = invemail.charAt(i);
        if (emailfind == '@')
        {
            emailfound = true;
        }

        else
        {
            enotfound = false;
        }


    }

edited Aug 30, 2015 at 16:36

answered Aug 30, 2015 at 15:15

Bruno Caceiro

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4 Comments

TNT Over a year ago

It seems like this will generate a variable emailfound may not have been initialized error from the compiler.

fabian Over a year ago

You need to initialize the variable too. Otherwise the read access in the condition of the if after the loop will cause a compile time error.

Adan Over a year ago

boolean emailfound = false;

Bruno Caceiro Over a year ago

Thanks, I missed the variable initialization. Fixed now.

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