Classes included within main() method - c++

If I have some code like
main(int argc, char *argv[])
{
...
#include "Class1.H"
#include "Class2.H"
...
}
Generally the main() method is the starting point of every application and the content within main() is to be executed. Am I right in the assumption that the content of all classes included into main() will be executed when main() is started?
greetings
Streight

No, no, NO.
First of all, you don't #include a file within a function. You #include a file at the beginning of a file, before other declarations. OK, you can use #include anywhere, but you really just shouldn't.
Second, #include doesn't execute anything. It's basically just a copy-paste operation. The contents of the #included file are (effectively) inserted exactly where you put the #include.
Third, if you're going to learn to program in C++, please consider picking up one of our recommended texts.
You commented:
I am working with the multiphaseEulerFoam Solver in OpenFoam and
inside the main() of multiphaseEulerFoam.C are classes included. I
assume that the classes have the right structure to be called in
main()
That may be the case, and I don't doubt that the classes have the right structure to be called from main. The problem is main will be malformed after the #includes because you'll have local class definitions and who knows what else within main.
Consider this. If you have a header:
foo.h
#ifndef FOO_H
#define FOO_H
class Foo
{
public:
Foo (const std::string& val)
:
mVal (val)
{
}
private:
std::string mVal;
};
#endif
And you try to include this in main:
main.cpp
int main()
{
#include "foo.h"
}
After preprocessing the #include directive, the resulting file that the compiler will try to compile will look like this:
preprocessed main.cpp
int main()
{
#ifndef FOO_H
#define FOO_H
class Foo
{
public:
Foo (const std::string& val)
:
mVal (val)
{
}
private:
std::string mVal;
};
#endif
}
This is all kinds of wrong. One, you can't declare local classes like this. Two, Foo won't be "executed", as you seem to assume.
main.cpp should look like this instead:
#include "foo.h"
int main()
{
}

#define and #include are just textual operations that take place during the 'preprocessing' phase of compilation, which is technically an optional phase. So you can mix and match them in all sorts of ways and as long as your preprocessor syntax is correct it will work.
However if you do redefine macros with #undef your code will be hard to follow because the same text could have different meanings in different places in the code.
For custom types typedef is much preferred where possible because you can still benefit from the type checking mechanism of the compiler and it is less error-prone because it is much less likely than #define macros to have unexpected side-effects on surrounding code.
Jim Blacklers Answer # #include inside the main () function

Try to avoid code like this. #include directive inserts contents of the file in its place.
You can simulate the result of your code by copy-pasting file content from Class1.H and Class2.H inside the main function.

Includes do not belong into any function or class method body, this is not a good idea to do.
No code will be executed unless you instantiate one of your classes in your header files.
Code is executed when:
Class is instantiated, then it's constructor method is called and the code inside the method is executed.
If there are variables of a class type inside your instantiated class, they will too run their constructors.
When you call a class method.
Try this example:
#include <iostream>
using namespace std;
int main()
{
class A
{ public:
A() { cout << "A constructor called" << endl; }
};
// A has no instances
class B
{ public:
B() { cout << "B constructor called" << endl; }
void test() { cout << "B test called" << endl; }
} bbb;
// bbb will be new class instance of B
bbb.test(); // example call of test method of bbb instance
B ccc; // another class instance of B
ccc.test(); // another call, this time of ccc instance
}
When you run it, you'll observe that:
there will be no instance of class A created. Nothing will be run from class A.
if you intantiate bbb and ccc, their constructors will be run. To run any other code you must first make a method, for example test and then call it.

This is an openFoam syntax he is correct in saying that open Foam treats #include like calling a function. In OpenFoam using #include Foo.H would run through the code not the class declaration that is done in a different hierarchy level. I would recommend all openFoam related question not be asked in a C++ forum because there is so much stuff built onto C++ in openFoam a lot the rules need to be broken to produce a working code.

You're only including declarations of classes. To execute their code, you need to create class instances (objects).
Also, you shouldn't write #include inside a function or a class method. More often than not it won't compile.

Related

C++ - how to forward declare a class in the same file as main() - getting variable uses undefined class error

I'm aware of using function prototypes, and I was under the impression that forward class declarations could serve a similar purpose when main() and a class are in the same file. For example, I would have expected this would compile:
// main.cpp
#include <iostream>
// class prototypes
class MyClass;
int main(void)
{
MyClass myClass;
// do stuff with myClass here
return(0);
}
class MyClass
{
public:
int someInt;
double someDouble;
// more stuff here . . .
};
But on the MyClass myClass; line I'm getting the error 'myClass' uses undefined class 'MyClass'. What am I doing wrong?
P.S. I'm aware that I could cut/paste main() below all the classes it uses and that would fix the error, but I'd prefer to keep main() as the first function or class.
P.P.S. I'm aware that in any substantial size production program main(), .h content, and .cpp content would be in 3 separate files. In this case I'm attempting to write a small example or test program where main and a class(es) are in the same file.
Forward declarations can only be used via pointers or references.
Calling a constructor function doesn't fall into this category.
I'm aware that I could cut/paste main() below all the classes it uses and that would fix the error, but I'd prefer to keep main() as the first function or class.
That's why usually header files are used, instead of placing all the declarations and definitions in the main.cpp file.
I'm aware that in any substantial size production program main(), .h content, and .cpp content would be in 3 separate files. In this case I'm attempting to write a small example or test program where main and a class(es) are in the same file.
You should still stick to that idiom though, everything else would probably end up in a mess.
This doesn't use forward declarations but it partially addresses the spirit of a single main.cpp with your "main" at the top. I find this technique sometimes useful when you want to share something via an online C++ ide where a single file is much easier to deal with, and you want to focus on the action in main rather than implementation detail in helper structs/classes etc.
#include <iostream>
template<typename MyClass,typename MyOtherClass>
int main_()
{
MyClass a;
a.do_foo();
MyOtherClass b;
b.do_bar();
return 0;
}
struct MyClass
{
void do_foo() { std::cout << "MyClass: do_foo called\n"; }
};
struct MyOtherClass
{
void do_bar() { std::cout << "MyOtherClass: do_bar called\n"; }
};
int main()
{
return main_<MyClass,MyOtherClass>();
}

Class redefinition: How to wrap methods within class [duplicate]

Is there a way to avoid the Graph:: repetition in the implementation file, yet still split the class into header + implementation? Such as in:
Header File:
#ifndef Graph_H
#define Graph_H
class Graph {
public:
Graph(int n);
void printGraph();
void addEdge();
void removeEdge();
};
#endif
Implementation File:
Graph::Graph(int n){}
void Graph::printGraph(){}
void Graph::addEdge(){}
void Graph::removeEdge(){}
I'm guessing this is to avoid lots of "unnecessary typing". Sadly there's no way to get rid of the scope (as many other answers have told you) however what I do personally is get the class defined with all my function prototypes in nice rows, then copy/paste into the implementation file then ctrl-c your ClassName:: on the clip board and run up the line with ctrl-v.
If you want to avoid typing the "Graph::" in front of the printGraph, addEdge etc., then the answer is "no", unfortunately. The "partial class" feature similar to C# is not accessible in C++ and the name of any class (like "Graph") is not a namespace, it's a scope.
No there's not. Not directly at least. You could go for preprocessor tricks, but don't do it.
#define IMPL Graph::
IMPL Graph(int n){}
void IMPL printGraph(){}
void IMPL addEdge(){}
void IMPL removeEdge(){}
Also, you shouldn't even want to do it. What's the point. Besides it being a C++ rule, it lets you know you're actually implementing a member function.
One option is using. If you have method definitions which are in a cpp file that never gets #included, then using is safe (doesn't affect other files):
foo.h:
class FooLongNameSpecialisationsParamaters
{
int x_;
public:
int Get () const;
void Set (int);
};
foo.cpp:
#include "foo.h"
using Foo = FooLongNameSpecialisationsParamaters;
int Foo::Get () const
{
return x_;
}
void Foo::Set (int x)
{
x_ = x;
}
main.cpp:
#include "foo.h"
int main ()
{
//Foo foo; <-- error
FooLongNameSpecialisationsParamaters foo;
return 0;
}
No, there is no way to avoid it. Otherwise, how would you know if a given function definition is for a class function or for a static function?
If you are asking if you can define a member function such as Graph::printGraph without specifying the class name qualification, then the answer is no, not the way that you want. This is not possible in C++:
implementation file:
void printEdge(){};
The above will compile just fine, but it won't do what you want. It won't define the member function by the same name within the Graph class. Rather, it will declare and define a new free function called printEdge.
This is good and proper, if by your point of view a bit of a pain, because you just might want two functions with the same name but in different scopes. Consider:
// Header File
class A
{
void foo();
};
class B
{
void foo();
};
void foo();
// Implementation File
void foo()
{
}
Which scope should the definition apply to? C++ does not restrict you from having different functions with the same names in different scopes, so you have to tell the compiler what function you're defining.
//yes it is possible using preprocessor like this:
#define $ ClassName //in .cpp
void $::Method1()
{
}
//or like this: in the header .h:
#undef $
#define $ ClassName'
// but you have to include the class header in last #include in your .cpp:
#include "truc.h"
#include "bidule.h" ...
#include "classname.h"
void $::Method() { }
//i was using also
#define $$ BaseClass
//with single inheritance than i can do this:
void $::Method()
{
$$::Method(); //call base class method
}
//but with a typedef defined into class like this it's better to do this:
class Derived : Base
{
typedef Base $$;
}
EDIT: I misread your question. This would be an answer to the question whether you can split header-files. It doesn't help you to avoid using LongClassName::-syntaxes, sorry.
The simple answer: You can split up c++-file, but you can not split up header-files.
The reason is quite simple. Whenever your compiler needs to compile a constructor, it needs to know exactly how many memory it needs to allocate for such an object.
For example:
class Foo {
double bar; //8 bytes
int goo; //4 bytes
}
new Foo() would require the allocation of 12 bytes memory. But if you were allowed to extend your class definitions over multiple files, and hence split header files, you could easily make a mess of this. Your compiler would never know if you already told it everything about the class, or whether you did not. Different places in your code could have different definitions of your class, leading to either segmentation faults or cryptic compiler errors.
For example:
h1.h:
class Foo {
double bar; // 8 bytes
int goo; // 4 bytes
}
h2.h:
#include "h1.h"
class Foo {
double goo; // 8 bytes
} // we extend foo with a double.
foo1.cpp:
#include "foo1.h"
Foo *makeFoo() {
return new Foo();
}
foo2.cpp:
#include "foo2.h"
void cleanupFoo(Foo *foo) {
delete foo;
}
foo1.h:
#include "h1.h"
Foo *makeFoo();
foo2.h:
#include "h1.h"
#include "h2.h"
void cleanupFoo(Foo *foo)
main.cpp:
#include foo1.h
#include foo2.h
void main() {
Foo *foo = makeFoo();
cleanupFoo(foo);
}
Carefully check what happens if you first compile main.cpp to main.o, then foo1.cpp to foo1.o and foo2.cpp to foo2.o, and finally link all of them together. This should compile, but the makeFoo() allocates something else then the cleanupFoo() deallocated.
So there you have it, feel free to split .cpp-files, but don't split up classes over header files.

How is it possible that I can make an instance from a class in which some member methods aren't defined yet? (C++)

I'm comparatively new to C++ so I tested some things out in Xcode, and found a really weird thing.
This is my 'Testing.h' file
#ifndef Testing_h
#define Testing_h
class Testing{
private:
int a;
public:
Testing(int a=3);
void hey(int b);
};
#endif
This is my 'Testing.cpp' file
#include "Testing.h"
Testing::Testing(int a){
a = 4;
}
And finally, this is the 'main.cpp' file
#include <iostream>
#include "Testing.h"
using namespace std;
int main(){
Testing a;
//Apparently not completing the definitions of every abstract methods in the class is not a problem
}
I only declared 'void hey(int b)' in 'Testing.h' but have not defined it in 'Testing.cpp'. So I was wondering how it is possible for the compiler to successfully compile the 'main.cpp' without having enough information of 'void hey(int b)'. Thanks in advance!
Because you never require there to be a definition for hey().
You can require a definition by calling it, for example :
a.hey(42);
And you'll see that the linker isn't too happy because hey is an undefined reference.
Testing a;//Apparently not completing the definitions of every abstract methods in the class is not a problem
You defined constructor with default value a=3 but calling both constructor argument and class parameter the same name is bad practice.
Instead you can write this:
//Testing.h
#ifndef Testing_h
#define Testing_h
using namespace std;
class Testing{
private:
int number;
public:
Testing(int a=3): number(a = 4){}//it's the same as your implementation in cpp file
void hey(int b);
int getNumber() {return number;}
};
#endif
//main.cpp
#include <iostream>
#include "Testing.h"
int main()
{
Testing object;
cout<<object.getNumber();// returns 4
return 0;
}
And why hey compiles?
During building your project compiler translates your source code into object code by verifying the syntax. After that process linker checks the definitions marked by whole phrases. Source code is compiled from each file provided. Linker doesn't care for the implementation presence, it only looks it up if a method is used by the program. So even without implementation of hey your program compiles.
Last remark
It's discouraged to include .cpp files use headers instead. Sometimes you can get yourself into multiple definitions of the same functions causing compiler errors.

While seperating classes, using a function in cpp file causes errors

So i just learned how to seperate classes and the youtube totourial is stressing on doing this alot, here's the link
https://www.youtube.com/watch?v=NTip15BHVZc&list=PLAE85DE8440AA6B83&index=15
My code is the exact same as his, and in the cpp file theres this thing:
mainClass::myfunction; (mainclass is the name of my class, myfunction is my function)
when i try to execute my program, it gives an error:
unidentified reference to 'mainClass::myfunction()'
here's my main.cpp file code:
#include <iostream>
#include "mainclass.h"
using namespace std;
int main()
{
mainClass bo;
bo.myfunction();
return 0;
}
here's my mainclass.h code:
#ifndef MAINCLASS_H
#define MAINCLASS_H
class mainClass
{
public:
myfunction();
};
#endif // MAINCLASS_H
my mainclass.cpp:
#include "mainclass.h"
#include <iostream>
using namespace std;
mainClass::myfunction()
{
cout << "I am a banana" << endl;
}
I don't know much about these so could you just tell me what the errors here are, because i copied everything correctly from the guy's totourial but still it doesn't work
P.S: this happens to me alot, i understand everything, nothing works, i copy everything, nothing works, and then i literally do exactly what the person is doing, still nothing works on all three of PC's, so i dont think the problem is with the devices lol
I doubt you completely copied and pasted that code because I'm fairly sure a teacher shouldn't be teaching having functions without a specified return type, but let's jump into it anyways...
Possibility #1
You meant to create a constructor for the class. In that case, please make sure the constructor function has the same name as the class. Also, you can't call it through .mainClass(), as it is a constructor.
class mainClass
{
public:
mainClass();
};
mainClass::mainClass()
{
cout << "I am a banana" << endl;
}
Possibility #2 You meant to create the class member function myfunction. You really should be specifying what return type your function is of. Some compilers will auto-assume int return type, and so the function you created is int myfunction();, but you really should be specifying it as void myfunction(); since you didn't return anything. Addl. info: Does C++ allow default return types for functions?
Next, change how you are giving the definition, by adding the return type.
void mainClass::myfunction()
{
cout << "I am a banana" << endl;
}
Possibility #3 Those should work, but another issue is that you might not have linked mainclass.cpp, so there is no definition available. In code blocks, right click on the project name and hit Add Files, then add the mainclass.cpp so the linker can define mainClass::myfunction().
To troubleshoot if the mainclass.cpp is being built with the project, try adding
#error I'm included! to the file mainclass.cpp after #include "mainclass.h". If you get an error I'm included!, then it is linked and you can remove the #error.

Is it possible to avoid repeating the class name in the implementation file?

Is there a way to avoid the Graph:: repetition in the implementation file, yet still split the class into header + implementation? Such as in:
Header File:
#ifndef Graph_H
#define Graph_H
class Graph {
public:
Graph(int n);
void printGraph();
void addEdge();
void removeEdge();
};
#endif
Implementation File:
Graph::Graph(int n){}
void Graph::printGraph(){}
void Graph::addEdge(){}
void Graph::removeEdge(){}
I'm guessing this is to avoid lots of "unnecessary typing". Sadly there's no way to get rid of the scope (as many other answers have told you) however what I do personally is get the class defined with all my function prototypes in nice rows, then copy/paste into the implementation file then ctrl-c your ClassName:: on the clip board and run up the line with ctrl-v.
If you want to avoid typing the "Graph::" in front of the printGraph, addEdge etc., then the answer is "no", unfortunately. The "partial class" feature similar to C# is not accessible in C++ and the name of any class (like "Graph") is not a namespace, it's a scope.
No there's not. Not directly at least. You could go for preprocessor tricks, but don't do it.
#define IMPL Graph::
IMPL Graph(int n){}
void IMPL printGraph(){}
void IMPL addEdge(){}
void IMPL removeEdge(){}
Also, you shouldn't even want to do it. What's the point. Besides it being a C++ rule, it lets you know you're actually implementing a member function.
One option is using. If you have method definitions which are in a cpp file that never gets #included, then using is safe (doesn't affect other files):
foo.h:
class FooLongNameSpecialisationsParamaters
{
int x_;
public:
int Get () const;
void Set (int);
};
foo.cpp:
#include "foo.h"
using Foo = FooLongNameSpecialisationsParamaters;
int Foo::Get () const
{
return x_;
}
void Foo::Set (int x)
{
x_ = x;
}
main.cpp:
#include "foo.h"
int main ()
{
//Foo foo; <-- error
FooLongNameSpecialisationsParamaters foo;
return 0;
}
No, there is no way to avoid it. Otherwise, how would you know if a given function definition is for a class function or for a static function?
If you are asking if you can define a member function such as Graph::printGraph without specifying the class name qualification, then the answer is no, not the way that you want. This is not possible in C++:
implementation file:
void printEdge(){};
The above will compile just fine, but it won't do what you want. It won't define the member function by the same name within the Graph class. Rather, it will declare and define a new free function called printEdge.
This is good and proper, if by your point of view a bit of a pain, because you just might want two functions with the same name but in different scopes. Consider:
// Header File
class A
{
void foo();
};
class B
{
void foo();
};
void foo();
// Implementation File
void foo()
{
}
Which scope should the definition apply to? C++ does not restrict you from having different functions with the same names in different scopes, so you have to tell the compiler what function you're defining.
//yes it is possible using preprocessor like this:
#define $ ClassName //in .cpp
void $::Method1()
{
}
//or like this: in the header .h:
#undef $
#define $ ClassName'
// but you have to include the class header in last #include in your .cpp:
#include "truc.h"
#include "bidule.h" ...
#include "classname.h"
void $::Method() { }
//i was using also
#define $$ BaseClass
//with single inheritance than i can do this:
void $::Method()
{
$$::Method(); //call base class method
}
//but with a typedef defined into class like this it's better to do this:
class Derived : Base
{
typedef Base $$;
}
EDIT: I misread your question. This would be an answer to the question whether you can split header-files. It doesn't help you to avoid using LongClassName::-syntaxes, sorry.
The simple answer: You can split up c++-file, but you can not split up header-files.
The reason is quite simple. Whenever your compiler needs to compile a constructor, it needs to know exactly how many memory it needs to allocate for such an object.
For example:
class Foo {
double bar; //8 bytes
int goo; //4 bytes
}
new Foo() would require the allocation of 12 bytes memory. But if you were allowed to extend your class definitions over multiple files, and hence split header files, you could easily make a mess of this. Your compiler would never know if you already told it everything about the class, or whether you did not. Different places in your code could have different definitions of your class, leading to either segmentation faults or cryptic compiler errors.
For example:
h1.h:
class Foo {
double bar; // 8 bytes
int goo; // 4 bytes
}
h2.h:
#include "h1.h"
class Foo {
double goo; // 8 bytes
} // we extend foo with a double.
foo1.cpp:
#include "foo1.h"
Foo *makeFoo() {
return new Foo();
}
foo2.cpp:
#include "foo2.h"
void cleanupFoo(Foo *foo) {
delete foo;
}
foo1.h:
#include "h1.h"
Foo *makeFoo();
foo2.h:
#include "h1.h"
#include "h2.h"
void cleanupFoo(Foo *foo)
main.cpp:
#include foo1.h
#include foo2.h
void main() {
Foo *foo = makeFoo();
cleanupFoo(foo);
}
Carefully check what happens if you first compile main.cpp to main.o, then foo1.cpp to foo1.o and foo2.cpp to foo2.o, and finally link all of them together. This should compile, but the makeFoo() allocates something else then the cleanupFoo() deallocated.
So there you have it, feel free to split .cpp-files, but don't split up classes over header files.