C++ Calling a function from the header file in the main - c++

I'm writing an integral calculator. I'm trying to get the main.cpp file to read in the functions from the functions class which is in the functions.h.cpp files. I have a function defined as the following for the easy case:
double square(double aX)
{
return aX*aX;
}
This function works when I include it in the main file, though it doesn't want to work when I try and call it from the main file when it's in the functions file. Here is the code that I have:
Main.cpp file:
#include <iostream>
#include "IntegralCalculator.h"
#include "functions.h"
using namespace std;
int main()
{
IntegralCalculator Function(square);
Function.print();
}
As you can see, there is another class called IntegralCalculator which actually does the calculation of the integral, though nothing should be wrong with that as it works when the function definition is included in the main, and the function is passed as a parameter. The functions.h file is defined as the following:
#ifndef FUNCTIONS_H
#define FUNCTIONS_H
#include <iostream>
#include <cmath>
#include "IntegralCalculator.h"
class functions : public IntegralCalculator
{
public:
functions();
double square(double aX);
};
#endif
and the functions.cpp file is defined as such:
#include <iostream>
#include <cmath>
#include "functions.h"
double functions::square(double aX)
{
return aX*aX;
}
I know that as of right now I don't need the <cmath> library, though I'm going to need it for more complicated functions when I add them.
I have tried inheriting the functions class in the main, and I have tried creating a namespace for the functions class and defining the function, however none of these have worked. When I go to compile, I get the following error:
Main.cpp: In function ‘int main()’:
Main.cpp:9: error: ‘square’ was not declared in this scope
If someone could please help me figure out this problem. This code that I'm trying to write is more advanced than my class has gone over, so I'm not sure how to fix it, though I want to write the program this way.
Edit: In case you were curious, I am including the functions.h and functions.cpp files in the Makefile. I was getting the same error when I was trying to put the square(double aX) definition in the IntegralCalculator class as well.

Your problem is, square is a member function AKA method. In other words, there is no plain symbol square anywhere, it's functions::square. And because it is not a static method, it makes no sense alone, it always needs a corresponding object instance, which you do not have and which IntegralCalculator couldn't use even if you did.
3 solutions, in the order of simplest, closest to your question, best:
From the constructor delcaration, move the square out of the class, make it a global function:
In functions.h outside any class:
double square(double aX);
In functions.cpp
double square(double aX) {
return aX*aX;
}
Usage in main.cpp:
IntegralCalculator Function(square);
Or make it static method (which is how you would have to do if you used a language like Java, which didn't allow "naked" functions, but is not really C++ way):
in functions.h:
class functions // inheriting IntegralCalculator removed as not needed
{
public:
static double square(double aX);
};
and in functions.cpp:
//static
double functions::square(double aX) {
return aX*aX;
}
Then you can pass it like this to `Function constructor in main.cpp:
IntegralCalculator Function(functions::square);
Finally, perhaps the best solution, sort of combining the best of the two above: put square in a namespace. So functions.h becomes
namespace functions {
double square(double aX)
}
functions.cpp becomes
namespace functions {
double square(double aX) {
return aX*aX;
}
}
You still use it in main like this:
IntegralCalculator Function(functions::square);

You have:
class functions : public IntegralCalculator
{
public:
functions();
double square(double aX);
};
which is ok, but doesn't tell other cpp files that there is a global function called square somewhere in your functions.cpp file.
What you need to do is include the header for that function so that the compiler knows that it exists:
double square(double ax);
class functions : public IntegralCalculator
{
public:
functions();
double square(double aX);
};
Also completely unrelated to your question, but worth mentioning after seeing your constructor code, you might want to use a typedef to declare the function pointer like so:
typedef double (IntegralFunctor*)(double);
So that your constructor can now be
IntegralCalculator::IntegralCalculator(IntegralFunctor func);
It's cleaner and less confusing this way

Related

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.

Getting error: expected constructor, destructor, or type conversion before ‘(’ token

I am trying to make functions repository. I have created four files:
Function.hpp, Function.cpp, FunctionsRepository.hpp, FunctionsRepository.cpp
I want to keep pointers to functions in vector of pointers.
//FunctionsRepository.hpp
#ifndef FUNCTIONSREPOSITORY_HPP
#define FUNCTIONSREPOSITORY_HPP
#include <vector>
using namespace std;
class FunctionsRepository {
private:
static vector<double *> pointerToFunctions;
public:
static void addFunction(double * wsk);
};
#endif
//FunctionRepository.cpp
#include "FunctionsRepository.hpp"
void FunctionsRepository::addFunction(double * wsk) {
pointerToFunctions.push_back(wsk);
}
//Functions.hpp
#ifndef FUNCTIONS_HPP
#define FUNCTOINS_HPP
#include "FunctionsRepository.hpp"
int constFunction(int numberOfVehicles);
void linearFunction();
void stepFunction();
#endif
//Funcctions.cpp
#include "Functions.hpp"
double constFunction(double numberOfVehicles){
return numberOfVehicles/2;
}
double (*funcConstant)(double) = constFunction;
//ERROR HERE
FunctionsRepository::addFunction(funcConstant);
I want to add new functions to program as easily as its possible and use it leater in other parts of program.
But I dont get it. Why i am getting this error. The addFunction() method is static, that means I can use it in other classes or parts of program. Vector is static to make sure that is the only one copy for whole program.
Use function wrapper. std::function can stores callable objects. So, your code will contain something like this:
class FunctionsRepository {
private:
// void() - function prototype
static std::vector<std::function<void()>> pointerToFunctions;
public:
static void addFunction(std::function<void()> wsk)
{
pointerToFunctions.push_back(wsk);
}
};
for more information consult official documentation: http://en.cppreference.com/w/cpp/utility/functional/function
I solved It. I received an error because I was calling the FunctionsRepository::addFunction(funcConstant); expression out of any scope. I just created new function to execute this command and thats all.

expected class-name before "{" token

So far I've looked up most of the typical solutions for cases of this error and yet non of them seems to work with my code. I'm using dev-c++
The structure of problematic class is following.
m.hh
#ifndef M_H
#define M_H
#include "z.hh"
#include <iostream>
#include <string>
using namespace std;
class M: public Z
{ /* this line is marked by the compiler as errorneous */
void m1();
void m2();
};
#endif
m.cpp
#include <iostream>
#include <string>
#include "m.hh"
using namespace std;
void M::m1() {/*bla bla*/};
void M::m2() {/*bla bla*/};
}
EDIT:
z.hh
#ifndef Z_H
#define Z_H
#include "m.hh"
#include <iostream>
#include <string>
using namespace std;
class Z;
static Z* s(string g, string i);
#endif
z.cpp
#include "z.hh"
#include <iostream>
#include <string>
class Z
{
public:
string i;
string g;
void set_i(string im) {i = im;}
string get_i() {return i;}
string get_g() {return g;}
virtual void m1()=0;
virtual void m2()=0;
Z* s(string g, string i) {
Z * z;
if(g=="m"){
M * z = new M;
}
}
};
Thanks!
Remove #include "m.hh" from z.hh to fix the circular include.
THE PROBLEM
You cannot derive from an incomplete type, which is what Z is in m.hh.
Upon deriving from a type the compiler must know the definition of said type. Just imagine you are trying to call a derived member function from Z on an object of type M, how would the compiler know if such call is a valid construct without knowing if Z actually declares such member?
And, a more relevant point, how would M be able to initialize Z without knowing what Z is?
THE SOLUTION
Move the definition of class Z to z.hh, and leave the definitions of its member functions in z.cpp (the same way you have split up M across m.hh, and m.cpp).
Well, the z.cpp file is actually a class declaration and should be a header file. A forward decalaration is only sufficient when you deal with pointers to that class, but not for inheriting from one (the compiler does need to know z's class layout for that).
The z.hh is not really needed, and the function declaration in it is not what you think (it's not a member function declaration). You can delete it and rename z.cpp to z.hh.
Ok, the exchange with #Luchian Grigore made me realize what you probably wanted and why you had "two z.hh". Z is to produce an M, but M inherits from Z, so there is a circular dependency. In C/C++ the way to break such a circle is to limit what one of the classes needs to know from the other: In order to declare a member function s that produces a pointer to a new M Z's class declaration only needs a typename M, not a the whole class declaration. So just a forward declaration (class M;) that introduces the type name M into the declaration of Z is enough; if the pointer to the M object is returned as a pointer to the base, Z, M's name is not even needed at all when Z is declared.
The definition of the factory function (be it file static or a member function) needs, if it is to produce an M, M's complete declaration though. That would be in a seaparate cpp file, which would include a complete header with M's full declaration.

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.