This question already has answers here:
Package accessibility for function and/or class
(6 answers)
Closed 9 years ago.
I'm wondering about how is it possible to set the package level access for a method in cpp.
Considering the following Java class
package Foo;
class Bar{
public Bar(){}
void bas(){}//package access, how in cpp?
}
the equivalent for cpp would be
namespace foo{
class bar{
public: bar(){};
//so now how would I set the package access specifier for bas?
};
};
As far as I know C++ has not package access level modifiers. But maybe friends class will be usefull in your case?
Related
This question already has answers here:
What "CV_EXPORTS_W" means
(2 answers)
Closed 5 years ago.
I'm trying to understand functionality of OpenCv library. I saw following strange type class declaration code here.
class CV_EXPORTS_W Subdiv2D
{
// code here
};
I don't understand, What does Subdiv2D do with class?
The Subdiv2D is the name of the class. The CV_EXPORTS_W part is probably some macro or an alias specific to Windows implementation.
This question already has answers here:
What are the differences between struct and class in C++?
(30 answers)
Closed 8 years ago.
I know that is possible for a struct:
struct A{
A* p;
};
But could we do that to a class:
class B{
B*p;
}
I tried to google it, but end not found it. Thank you very much!
Yes. There is no difference between a class declared with struct and one declared with class other than the default access for members and bases.
Yes, a class can contain pointers to a class of the same type, and the symbol representing the class can appear in signatures of functions or in the bodies of functions defined in that class. Furthermore, it should be noted that, in C++, struct and class are identical with only one difference -- the default access level for a struct is "public", while the default access level for a class is "private" (of course, one can trivially write "public:" or "private:" at the beginning of either in order to change the access level; it is only convention that results in "class" being used for types that include more functionality, encapsulation and "struct" for pure data objects).
This question already has answers here:
How do you declare an interface in C++?
(17 answers)
Closed 9 years ago.
I want to know about the way of defining interface class in C++ by this approach. Is it possible? I don't want to know about interface creation but want to know about the abstract keyword use in C++.
In C++, an abstract class is any class that has at least 1 pure virtual function.
C++ does not have direct support for interfaces, but you can make one by making all of the functions public, virtual, and abstract, and have no data members in the class.
In C++, an interface is defined as follows:
class Interface {
public:
virtual ~Interface();
virtual void aMethod() = 0;
};
Note the virtual destructor.
This question already has answers here:
When do I need anonymous class in C++?
(6 answers)
Unnamed classes
(2 answers)
Closed 9 years ago.
I have been exploring more C++ and encountered a phenomenon like Unnamed classes.
class : public classB{
public:
protected:
private:
};
How are these classes useful? Is it like the concept of lambda functions?
I would like to know the purpose of such classes. The other threads don't really show any usefulness of the feature.
This question already has answers here:
Default class inheritance access
(4 answers)
Closed 3 years ago.
I have some legacy code that I have to wrap, and I have come across this declaration:
class Foo : Bar
{
// ...
};
This seems to compile under GCC. I know it's bad, but I can't change it. My question is, if no inheritance access specifier is present, how does the C++ compiler handle it?
For classes, the default is private.
For structs, the default is public.
BTW, it is not called access modifier. It is called access specifier
$11.2/2 - "In the absence of an
access-specifier for a base class,
public is assumed when the derived
class is defined with the class-key
struct and private is assumed when the
class is defined with the class-key
class."
In your context, 'Bar' is a private base class of 'Foo'
It's private.