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I am learning data structures and algorithms for C++ from Goodrich. They have given this LinkedList implementation. I understand the code but I am not but I am not able to use this in main class. How do I construct an instance and make insert, delete? For example I tried to create an instance of the class as follows:

StringLinkedList() L;

But it is showing the error: expected ";" before 'L

#include <iostream>
#include <string>

using namespace std;

class StringNode {
private:
    string elem;
    StringNode* next;

    friend class StringLinkedList;
};

class StringLinkedList{
public:
    StringLinkedList();
    ~StringLinkedList();
    bool empty() const;
    const string& front() const;
    void addFront(const string& e);
    void removeFront();
private:
    StringNode* head;
};

StringLinkedList::StringLinkedList()
    :head(NULL) {}
StringLinkedList::~StringLinkedList()
    {while (!empty()) removeFront();}
bool StringLinkedList::empty() const
    {return head==NULL;}
const string& StringLinkedList::front() const
    {return head->elem;}
void StringLinkedList::addFront(const string& e){
    StringNode* v = new StringNode;
    v->elem=e;
    v->next = head;
    head=v;
}
void StringLinkedList::removeFront(){
    StringNode* old=head;
    head = old->next;
    delete old;
}


int main () {
}
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    "but I am not but I am not able to use this in main class" - what is the exact problem you have with it? please show some example code of how you use it. Commented Aug 20, 2013 at 8:52
  • Ok. I tried to construct an instance of the class using constructor like this: StringLinkedList() L; But it is showing the error expected ";" before 'L'. I don't know any other way of creating the instance. Commented Aug 20, 2013 at 8:53
  • please edit the question when providing more details, don't add that information to the comments Commented Aug 20, 2013 at 8:54
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    This "linked list implementation" looks more like a stack implemented via linked list to me: You can only access the head of the list, at any time, and there aren't even any methods for iterating through it. Commented Aug 20, 2013 at 8:56
  • 1
    Do you know how to instantiate any class in C++? Maybe that is the problem. Commented Aug 20, 2013 at 8:59

3 Answers 3

Reset to default
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Brackets () indicate a function call. If you want to declare a variable the syntax is

Typename variable_name;

Optionally, you may need to pass parameters to a constructor

Typename variable_name(param);

In C++11 the uniform initialisation syntax allows you to use {} but I digress. Either way they come after the variable name. In your case, this works:

StringLinkedList L;


When you say 

StringLinkedList() L;

the compiler sees a typename, then expects a variable name, but gets () before the name L (BTW - it might deserve a longer name) so decided you must be making a function call, which should end with a semicolon. But it doesn't, it ends with L; so you get

expected ";" before 'L
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4 Comments

+1 Answer reduced to the real problem. Well explained I think.
+1 - It's great to see someone taking the time to answer a question that is clearly for a beginner, rather than complaining about them not understanding what they're trying to learn.
@Merlin069 many users of SO have the bad habit of attack this kind of noob questions as if they are vultures, instead of answering the cuestion realizing that a few time ago they were noobs too.
@Manu343726 exactly. Before entering SO, egos should be left at the door!
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You can create an instance and add and remove items like this:

int main () {
    StringLinkedList list; // construct an instance
    list.addFront("foo");  // Add "foo"
    list.addFront("bar");  // Add "bar"
    list.removeFront();    // Remove "bar"
    // List is automatically deleted now
}
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6 Comments

Thanks. Why doesn't it work if I add the brackets next to method name like StringLinkedList() L;?
@user1425223 Why should that work? It doesn't make sense syntactically.
@user1425223 you mean declaring the variable with the following sentence?: StringLinkedList() list;
@Manu343726 yes, that's what I meant.
It won't work because it's not valid syntax in C++. Why should it work?
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The method that you use for creation of object isn't correct i dont know what do you mean by

StringLinkedList() L;

the StringLinkedList() is a call for the constructor of the class and it can be used for creating objects of class but no need to specify a constructor call unless needed. You can use the constructor if you are creating object and also want to initialise data members of objects for its creation.

StringLinkedList L=StringLinkedList();

is the correct method but you need to write only 

StringLinkedList L;

as it will automatically call the default constructor. The actual process happening is the constructor create a temporary object and then assign it to your object variable L; but it will be done automatically

I dont see any problem in declaring the objects of linked list class like this.

int main ()
{
    StringLinkedList s1;
    s1.addFront("hai");
    s1.addFront("dude");
    s1.removeFront();
    cout<<s1.front();
}
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1 Comment

Note that StringLinkedList L = StringinkedList() is not the same as StringLinkedList L;. The first is a declaration of a linked list wich uses the copy constructor to initialize the variable with the temporal object created by the explicit call to the constructor in the right side. The second is a variable declaration wich simply calls the default ctor. On the other hand, any decent C++ compiler could elide the copy in the first case.

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