Avoid exposing static variable for a default function argument - c++

I have a function foo() that updates a global CAtlList. Now I want to reuse foo to update another list. Currently I achieve this using a default argument.
//header.h
extern CAtlList<data> globalList;
void foo(CAtlList<data> &somelist = globalList);
//file1.cpp
CAtlList<data> globalList;
void foo(CAtlList<data> &somelist)
{
//update somelist
}
//file2.cpp
#include "header.h"
foo();
and
CAtlList<data> anotherList;
foo(anotherList);
//use anotherList
But for the default scenario foo takes globalList by reference, which means globalList must be visible at the point of declaration. I had to expose and add an extern declaration of globalList for that to happen.
I'd rather not expose it, is it at all possible?

Use an overload, not a default parameter.
void foo();
void foo(CAtlList<data> &somelist);
It costs you 4 short lines of code in the .cpp: The body of the parameterless foo calling the 1-parameter foo with the correct argument. That's the total cost to get the expressiveness you desire. A default parameter is not the tool for your job.

I don't know you can avoid declaring a variable that's referenced as a default. However, you could simply overload the function:
//file1.h
void foo();
void foo(CAtlList<data> &somelist);
//file1.cpp
#include "file2.h"
CAtlList<data> globalList; // Local scope only
void foo(CAtlList<data> &somelist)
{
//update somelist
}
void foo()
{
foo(globalList);
}
A more modern way is to use the 'optional' class that's part of the C++ library:
http://en.cppreference.com/w/cpp/utility/optional
//file1.h
#include <optional>
void foo(optional<CAtlList<data>> &somelist);
//file1.cpp
#include "file1.h"
CAtlList<data> globalList; // Local scope only
void foo(optional<CAtlList<data>> &somelist){
if (somelist.has_value()) { /* update somelist */ }
else { /* update globalList */ }
}
However, I suspect that this will break the existing dependencies. If there are few and easy to fix, this is probably the way.
The other comments are also correct. You should try and avoid the use of global state wherever possible - its a symptom of poor design.

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.

Invoking a member method pointer from another method

In class Foo I have two methods, assign_handler() and call_handler().
The actual handler code is in the main.cpp which is do_this(). do_this() uses the some global variables in main.cpp,
I think Foo has to have a function pointer as member which will be assigned in assign_handler() which is what I did. However I'm having trouble invoking assign_handler() i.e. calling do_this(), from call_handler().
Note: call_handler() itself is call by a sigaction in Foo.
EDIT: I tried producing a MCVE as suggested in the comments. I've used gedit to create the files and compile it using g++ in command line. The code works. However in my Eclipse project I get the errors shown in inline comments of the code.
MCVE:
//Foo.h
class Foo{
public:
void (*funptr)(void);
void call_handler();
void assign_handler (void(*func1)(void));
Foo(){};
};
//Foo.cpp
#include "Foo.h"
void Foo::assign_handler(void(*func1)(void)){
funptr = func1;
}
void Foo::call_handler(){
funptr();//error: invalid use of member Foo::funptr in static member function; from this location
//or
//this->funptr();//error: 'this' is unavailable for static member functions
}
//main.cpp
#include <iostream>
#include "Foo.h"
using namespace std;
void do_this(void);
int main(void){
Foo foo;
foo.assign_handler(do_this);
foo.call_handler(); //this won't be called explicitly, it is assigned as a handler for a sigaction
int x;
cin>>x;
}
void do_this(void){
cout<<"done"<<endl;
}
I'll divide my answer in two parts. First I'll attempt to answer your question, then I'll attempt to tell you what you actually want to do.
Your question is how to assign a function pointer to a member variable and then call it from a static member function. Since the function pointer is a member of the class you will also require a pointer to the class in order to call the function pointer. A way of achieving this is to add a static member to your class that holds a pointer to the (single) instance of your class. Since you indicated that you will be using this as a signal handler, you won't want to use multiple handlers anyway.
So, something like this:
//Foo.h
class Foo{
public:
static void call_handler();
void assign_handler (void(*func1)(void));
Foo() {
ms_instance = this;
};
private:
void (*funptr)(void);
static Foo *ms_instance;
};
//Foo.cpp
#include "Foo.h"
void Foo::assign_handler(void(*func1)(void)){
funptr = func1;
}
void Foo::call_handler(){
ms_instance->funptr();
}
A more general way would be to store a function object:
//Foo.h
#include <functional>
#include <utility>
class Foo{
public:
static void call_handler();
template<typename func>
void assign_handler (func&& handler)
{
m_handler = std::forward(handler);
}
Foo() {
ms_instance = this;
};
private:
std::function<void(void)> m_handler;
static Foo *ms_instance;
};
//Foo.cpp
#include "Foo.h"
void Foo::call_handler(){
ms_instance->m_handler();
}
This way you can assign lots of different stuff as the handler:
// Function pointers
foo.assign_handler(do_this);
// Lambdas
foo.assign_handler([]() { /* do something */ });
// Binds - you should probably prefer lambdas...
foo.assign_handler(std::bind(&MyClass::member_func, &myObj));
Now what you actually want to do when you are going to handle a signal is a bit more complicated. Remember that signal handlers can only call certain functions (async-signal-safe functions) - otherwise things may get ugly. Therefore there is a common trick that you should perform called the self pipe trick. Essentially you should have a signal handler that receives the signal, but only calls write on a pipe with the signal number as the data to send. Then you have another place in your code that calls select on the pipe and then read to read the signal number. You then call the appropriate handler function which is then allowed to do whatever you like.
An example of this is here: http://man7.org/tlpi/code/online/book/altio/self_pipe.c.html
Be aware that it can be slightly tricky to get this right in a cross-platform manner, especially if multithreaded.

Use case of C++11 function declaration at block scope?

It would seem I can declare a function at block scope:
int main()
{
void f(); // OK
}
However I can't define it:
int main()
{
void f() {}; // ERROR
}
My question is, of what use is a function declaration at block scope? What is a use case?
It's sometimes a shortcut to declaring and calling an externally-linked function which itself isn't publically defined in a header. For example, imagine you were linking against a C library which you knew provided a LibDebugOn(int) call but hadn't defined it in a header. It can be a shortcut to declare and call it in one place:
void myFunc() {
// call "Lib" but turn on debugging via hidden API
extern "C" void LibDebugOn(int); // declare hidden C-linked function
LibDebugOn(1); // call it
// do something with the library here...
LibDebugOn(0); // turn off lib debugging now
}
In fairness this is usually only worthwhile for a one-off quick hack, and not something to be encouraged.
You can define it. http://ideone.com/kJHGoF
#include <cstdio>
int main()
{
void f(); // Forward declare function named "f" returning void in the
// global namespace.
f();
}
/*
void g()
{
f(); // ERROR!
}
*/
void f()
{
std::puts("hello!");
}
I'm not sure why someone would actually want to use this. It is in the language this way for backwards compatibility with C; but I've got no idea what someone would do with this in C.

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.

C++ accessing a function from in a class, receiving functions as a parameter

i have two questions that are fairly small and related so i will put them both in the same question.
i have been experimenting with classes and i was attempting to access a class in another file that wasn't in a class so for example.
//class 1 .cpp
void Class1::function1()//another error
{
function()
}
//main.cpp
void function()
{
//stuff happens
}
is there a way to-do this? or would i need to add this function to a class to get it to work. also how would you go about creating a function that receives a function as it parimetre? for example function(function2())
i am simply trying to access a function from a class as it would make my code easier to use later if the function that i am using doesn't get added to a class. with regards to the seconds question i which to create a function that receives a time and a function as an argument. it will wait for the specified time then execute the program
How to access a function in another file?
Depends on the type of function, there can be to cases:
1. Accessing class member functions in another file(Translation Unit):
Obviously, you need to include the header file, which has the class declaration in your caller translation unit.
Example code:
//MyClass.h
class MyClass
{
//Note that access specifier
public:
void doSomething()
{
//Do something meaningful here
}
};
#include"MyClass.h" //Include the header here
//Your another cpp file
int main()
{
MyClass obj;
obj.doSomething();
return 0;
}
2. Accessing free functions in another file(Translation Unit):
You do not need to include the function in any class, just include the header file which declares the function and then use it in your translation unit.
Example Code:
//YourInclude.h
inline void doSomething() //See why inline in #Ben Voight's comments
{
//Something that is interesting hopefully
}
//Your another file
#include"YourInclude.h"
int main()
{
doSomething()
return 0;
}
Another case as pointed out by #Ben in comments can be:
A declaration in the header file, followed by a definition in just one translation unit
Example Code:
//Yourinclude File
void doSomething(); //declares the function
//Your another file
include"Yourinclude"
void doSomething() //Defines the function
{
//Something interesting
}
int main()
{
doSomething();
return 0;
}
Alternately, a messy way to do this can be to just mark the function as extern in your another file and use the function.Not recommended but a possibility so here is how:
Example Code:
extern void doSomething();
int main()
{
doSomething();
return 0;
}
How would you go about creating a function that receives a function as it parameter?
By using function pointers
In a nutshell Function pointers are nothing but pointers but ones which hold address of functions.
Example Code:
int someFunction(int i)
{
//some functionality
}
int (*myfunc)(int) = &someFunction;
void doSomething(myfunc *ptr)
{
(*ptr)(10); //Calls the pointed function
}
You need a prototype for the function you want to call. A class body contains prototypes for all its member functions, but standalone functions can also have prototypes. Typically you organize these in a header file, included from both the file which contains the function implementation (so the compiler can check the signature) and in any files which wish to call the function.
(1) How can the `class` function be accessible ?
You need to declare the class body in a header file and #include that wherever needed. For example,
//class.h
class Class1 {
public: void function1 (); // define this function in class.cpp
};
Now #include this into main.cpp
#include"class.h"
You can use function1 inside main.cpp.
(2) How to pass a function of class as parameter to another function ?
You can use pointer to class member functions.