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initialization of 'element' is skipped by 'case' label [duplicate]

Tags:

c++

debugging

I don't understand why I am getting the error:

initialization of 'element' is skipped by 'case' label.

Can someone please explain to me?

void LinkedList::process_example(int choice) {
    switch(choice) {
    case 1:
        cout << endl << endl << "Current S = ";
        this->printSet();

        cout << "Enter an element :";
        char* element = "lol";

        //cin>>element;
        cin.clear();
        cin.ignore(200, '\n');

        this->Addelementfromback(element); //error is here
        cout << endl << endl << "Current S = ";

        this->printSet();
        break;

    case 2:
        this->check_element();
        break;

    case 3:
        cout << endl << endl;
        cout << "Current Set S = ";
        this->printSet();

        cout << endl << "S has ";
        int count = this ->check_cardinality();
        cout << count << " elements";
        break;
    }
}
like image 594
Computernerd Avatar asked Feb 03 '13 04:02

Computernerd


2 Answers

Try wrap case with {}, and put all your statement inside {}.

case 1:
{
   cout << endl << endl << "Current S = ";
   this->printSet();    
   // and other mess
}
break;

You should put all these statement in functions, keep case statement clear. For example, write this style:

case 1:
   initializeElement();
   break;
case 2:
   doSomethingElse();
   break;

See link

like image 82
billz Avatar answered Nov 16 '22 20:11

billz


When a variable is declared in one case, the next case is technically still in the same scope so you could reference it there but if you hit that case without hitting this one first you would end up calling an uninitialised variable. This error prevents that.

All you need to do is either define it before the switch statement or use curly braces { } to make sure it goes out of scope before exiting a specific case.

like image 29
Rhys Thompson Avatar answered Nov 16 '22 19:11

Rhys Thompson