Error - "expression must have a class type" - c++

Question

I have my own LinkedList class, and when I'm creating an instance in the main i get the "expression must have class type" error. I have a default c'tor in the LinkedList class.

so when I'm trying to do this :

    LinkedList<Animal> aL();

    for (int i = 0; i < numOfAnimals; i++)
    {

        aL.addLast(*animals[i]);
    }

    cout << aL << endl;

it won't compile. but if I declare like this:

    LinkedList<Animal> aL = LinkedList<Animal>();

it works. of course that the secont option isn't efficient, and i rather use the first one. can anyone explain me why it doesn't work or how to fix it? Thnaks!

`LinkedList aL();` -- That is a function declaration. Look closely at it. — PaulMcKenzie, Oct 08 '17 at 13:23
@VittorioRomeo As far as I know, the term "most vexing parse" refers to a bit more involved variation of this, like `A a(B());` — lisyarus, Oct 08 '17 at 13:23
@lisyarus - Considering it's not an official term, but a colloquial one, I don't see why not expand it. It's all very vexing if one isn't used to it. — StoryTeller - Unslander Monica, Oct 08 '17 at 13:27
@StoryTeller Doesn't matter whether the term is official or not. It already *has* a defined meaning in C++ world. If one uses the term to mean something different, he may come into trouble. I agree that this could also be called "most vexing parse", since it is rather confusing sometimes, but still: the term already has a meaning. — lisyarus, Oct 08 '17 at 13:31
@lisyarus - No trouble. Colloquialisms shift often, and with use. — StoryTeller - Unslander Monica, Oct 08 '17 at 13:34

lisyarus · Accepted Answer · 2017-10-08T13:33:40.540

The line

LinkedList<Animal> aL();

is not creating an instance of a LinkedList<Animal>, but is declaring a function aL with no arguments and LinkedList<Animal> as its return type.

To create an object using the default constructor, just don't use any braces:

LinkedList<Animal> aL;

Alternatively, if you wish to be that explicit, you can use curly braces from C++11

LinkedList<Animal> aL{};

Or, if you like function call syntax that much, you can

LinkedList<Animal> aL = LinkedList<Animal>();

or even

auto aL = LinkedList<Animal>();

Error - "expression must have a class type" - c++

1 Answers1