Error: Expression Must Have a Class Type

Expression must have class type

It's a pointer, so instead try:

a->f();

Basically the operator . (used to access an object's fields and methods) is used on objects and references, so:

A a;
a.f();
A& ref = a;
ref.f();

If you have a pointer type, you have to dereference it first to obtain a reference:

A* ptr = new A();
(*ptr).f();
ptr->f();

The a->b notation is usually just a shorthand for (*a).b.

A note on smart pointers

The operator-> can be overloaded, which is notably used by smart pointers. When you're using smart pointers, then you also use -> to refer to the pointed object:

auto ptr = make_unique<A>();
ptr->f();

Why am I getting expression must have class type error?

You are accessing the pointer incorrectly. Accessing inner variables from a pointer require -> instead of the . operator.

Switch your code:

Student* student_ptr = new Student;
student_ptr->name = "Tom";
student_ptr->id = 1;

Alternatively, if you really want to use the . operator, you can do:

Student* student_ptr = new Student;
(*student_ptr).name = "Tom";
(*student_ptr).id = 1; // the more you know

error: expression must have a class type

Most vexing parse :

This line:

Worker myWorker();

declares a function taking no parameters and returning a Worker.

Simply declare your object with :

Worker myWorker;

What is Expression must have class type

It's probably because this is a pointer and to try to access it using ..
Moreover malloc returns a void* so you must cast

Try writing this instead :

this->ptr = static_cast<int *>(std::malloc(4));

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

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>();

Expression must have a class type - C++/CLI

The error message is a bit misleading, in a way, because it implies that the problem is with the actual listviewX and listviewY variables (which, as you have pointed out, are 'hatted' class types, or handles).

However, the SubItems member of the ListViewItem class is an array of class handles (not raw classes), so you need the -> operator on the two occurrences of that, too (rather than the . operator).

This is what you need the code to be:

compareResult = ObjectCompare->Compare(listviewX->SubItems[ColumnToSort]->Text,
        listviewY->SubItems[ColumnToSort]->Text);

Expression must have class type but it has type *shape

If input_shape_pointer is a pointer to a shape then use -> instead of .. For example, the expression

input_shape_pointer.get_dimensions()

should be replaced by:

input_shape_pointer->get_dimensions()

Note the use of -> instead of . in the above expression

The expression must have a class type

Since C++17 (and to a lesser extent 14) we can use std::filesystem (std::experimental::filesystem in C++14) to manipulate files and create directories.

For example in your case:

...
std::filesystem::path DDDirectory("C:\\dnGames\\DuelistsDance"); 

try {
  std::filesystem::create_directories(DDDirectory); // Creates all the directories needed, like mkdir -p on linux

// Success here

} catch(std::filesystem::filesystem_error& e) {
  // Handle errors here
}

This will make handling of your errors cleaner and your code cross-platform (although you will have to change the path, but std::filesystem::path turns / into \\ on windows anyway). It also makes your code easier to read and as you can see, much much shorter.

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