How to implement a #define in another #define? - c++

The previous problem description was ambiguous, so I modified something below. Thanks.
I want to implement some macros like this:
#define AddVariable(x) \
#define x (++counter)
class Base {
private:
int counter = 0;
}
class Extend : public Base {
public:
void Handler() {
AddVariable(ASDF);
AddVariable(HJKL);
// Here may add more variables.
}
void AnotherHandler() {
// It calls ASDF, HJKL too.
}
}
The ASDF and HJKL should be available through all handlers in the cpp file, so I have to define it in the macro (it's not a good design, though). But how should I write the proper macros to achieve it (#define cannot be nested in another #define)? Or is there another better way of implementation?
Thanks in advance.
Update:
A potential implementation is
#define AddVariable(x) \
int x = ++counter;
It works, but x is not global, and I have to fix this.

It looks like you are trying to expose an incrementer to a Base object's counter to all of your functions in the .cpp file.
Answer: This is not possible.
The other functions/objects in your .cpp file do not have a reference to a Base object and thereby cannot change any of it's data.
If you want to maintain a single counter for all the Base objects you might try something like this:
class Base {
public:
static void ASDF(){counter++;}
private:
static int counter = 0;
};
This can be called from and other function:
void AnotherHandler() {
Base::ASDF();
}
EDIT:
class Base {
protected:
static int counter = 0;
};
class Another : public Base{
public:
Another(){
counter++; // As a child of Base you have access to all of it's protected variables
}
}

Would using an std::map work?
#include <iostream>
#include <map>
#include <string>
class Base {
protected:
void AddVariable(const std::string& name) {
variables[name] = counter++;
}
int& GetVariable(const std::string& name) {
return variables[name];
}
private:
int counter = 0;
std::map<std::string, int> variables;
};
class Extend : Base {
public:
void Handler() {
AddVariable("ASDF");
AddVariable("HJKL");
// May add more variables here ...
}
void AnotherHandler() {
// Use ASDF, HJKL here too
std::cout << GetVariable("ASDF") << std::endl;
std::cout << GetVariable("HJKL") << std::endl;
}
};
int main()
{
Extend e;
e.Handler();
e.AnotherHandler();
}
Output:
0
1

You cannot do that (defining a macro in a macro) with the C preprocessor.
However, sometimes you could play X-macro tricks, like here.
If you really need macro defining macros, switch to some external or non-standard preprocessor, like GNU m4 or gpp.
Alternatively, generate your C++ code with some external tool (e.g. your own Python or GNU awk script)
At last, recent GCC (e.g. GNU cpp) or Clang/LLVM offer the __COUNTER__ macro (also usable indirectly with stringification & concatenation)

Related

Compiling two different implementations of the same class

Currently I am writing a class that supports data proccessing on the cpu or gpu utilizing preprocessor definitions to determine which header file to include.
IE
#ifdef CPU_work
#include "cpu_backend.h"
#endif
#ifdef GPU_work
#include "gpu_backend.h"
#endif
class Work {
//Implementation dependant upon included header
}
However, there maybe instances where I would need both variants. Is there anyway I could do something like....
namespace CPU {
#define CPU_work
//Generate implementation of WorkClass with cpu_backend.h
}
namespace GPU {
#define GPU_work
//Generate implementation of WorkClass with gpu_backend.h
}
and therefor determine which implementation I want via something like...
CPU::Work cpuObject;
GPU::Work gpuObject;
Would be happy with any work-arounds also.
Much thanks JJ.
This might be the place to use a template method design. Your base class implements everything that is common to both CPU and GPU and then you use abstract functions where there are differences.
class Work {
public:
void execute() {
// Do some initializing
foo();
// Do some middle stuff
bar();
// Do some final stuff
}
private:
virtual void foo() = 0;
virtual void bar() = 0;
}
class CpuWork: public Work {
virtual void foo() {
// Do some CPU stuff
}
virtual void bar() {
// Do some more CPU stuff
}
}
class GpuWork: public Work {
virtual void foo() {
// Do some GPU stuff
}
virtual void bar() {
// Do some more GPU stuff
}
}
You now can't use your base class Work by accident since it's abstract and you can't accidentally invoke your derived classes foo or bar since they are private members of the base class.
Interesting question:) If I understood your goals correct, I can suggest a few solutions.
First uses template specialization, template default arguments and (of course) some macros.
Check this out:
// cpu_backend.h
#define CPU_BACKEND
class UseCPU;
#ifndef GPU_BACKEND
template<class Method = UseCPU>
struct Backend;
#endif
template<>
struct Backend<UseCPU>
{
char* Info() { return "CPU"; }
};
// gpu_backend.h
#define GPU_BACKEND
class UseGPU;
#ifndef CPU_BACKEND
template<class Method = UseGPU>
struct Backend;
#endif
template<>
struct Backend<UseGPU>
{
char* Info() { return "GPU"; }
};
// main.cpp
// Try to swap comments on headers
// and see how output changes
#include "cpu_backend.h"
//#include "gpu_backend.h"
#include <iostream>
template<class ... Method>
struct Work
{
Work()
{
std::cout << "I use " << backend.Info() << std::endl;
}
private:
Backend<Method ...> backend;
};
int main()
{
Work<> work;
// Uncomment these two while including both headers
//Work<UseCPU> cpuWork;
//Work<UseGPU> gpuWork;
return 0;
}
If you use MSVC you can simplify example above eliminating #define and #ifndef.
Trick: MSVC (2017 and maybe earlier versions) allow to omit that macros thresh, just ignoring the second declaration if they meet in
the same compilation unit, like this:
template<class Method = UseCPU>
struct Backend;
template<class Method = UseGPU>
struct Backend;
BUT this will be not standard. Standard does not allow specifying default template args twice.
Meanwhile, this solution has few drawback:
When you include both headers, someone still can say Work<> which will
use the backend specified by the first header you included.
However, it would be better if compiler forced a person to specify a
backend type explicitly in this circumstances, because otherwise it
relies on the header inclusion order which is bad (say hello to
macros).
Also, it assumes that both backends have the same API (like Info()
in my case)
Possible Fixes for those:
I am sure it is possible to make compiler give an error when both
headers are included and no explicit backend was specified, but it
probably involves more preprocessor things or some SFINAE...
If your backends do have different APIs, then you can insert a few
#ifdef where needed or (preferably) use C++17
if constexpr(std::is_same<Method, UseCPU>()::value) if you have access
to such cool features:)

template class with functions that do not use the template

I am jumping through hoops to reduce inheritance.
I read one similar question here. It shows how the issue can be resolved using a base class. I try to loose inheritance, so I am looking for something different - more along the lines of annotation.
I create and compile a template class with one specialisation (normal). The method that requires the template is in the header (Mixer.hpp). The method that does not require the template is in the cpp file (Mixer.cpp). When compiled into a static library, the cpp part only gets compiled as one specialisation (Mixer<normal>). The compiler does not know about (awsome) at that time. Importing the resulting static library into another project and attempting to create a different generic (awsome) class results in a linker error because obviously the library does not contain that method identifier (Mixer<awesome>::noTemplateInvolved). However the code for the normal implementation is as good as any so really the linker could just link to the existing source of the other template version (Mixer<?dontcare?>::noTemplateInvolved). All that the compiler has to do is to mark it appropriately for the linker.
Here is source code that results in a linker error:
//Target compiled to Provider.lib
//Mixer.hpp
#pragma once
#include <iostream>
using namespace std;
struct normal { static void log() { cout << "normal\n"; } };
template<typename output = normal>
class Mixer
{
public:
void callingTemplate();
void noTemplateInvolved();
};
template<typename output>
void Mixer<output>::callingTemplate() { output::log(); }
//Mixer.cpp
#include "Mixer.hpp"
void Mixer<>::noTemplateInvolved()
{
cout << "noTemplateInvolved\n";
}
//Target Compiled to User.exe
//This target imports Provider.lib
#include <Provider\Mixer.hpp>
#pragma comment(lib, "Provider.lib")
struct awsome { static void log() { cout << "awsome\n"; } };
int main()
{
Mixer<> n;
n.callingTemplate();
n.noTemplateInvolved();
Mixer<awsome> a;
a.callingTemplate();
a.noTemplateInvolved(); //linker error here
return 0;
}
The class Mixer<awsome> can link to the method callingTemplate because its definition is in the header and the compiler creates that function. At User.exe compile time the definition of noTemplateInvolved is hidden from the compiler. The compiler can not create that method and linking has to fail.
There are three solutions that I am aware of.
move the definition of noTemplateInvolved to the header.
include the cpp file
inherit from a baseclass
I am looking for another solution. The body of noTemplateInvolved really has nothing to do with the template. I would like to annotate the method in the header. I want the compiler to know it should always use the same base implementation regardless of the template.
Is that at all possible?
EDIT: Annotated that boring paragraph at the beginning a bit.
The answer turns out to be a base class as suggested in the comments. One of the reasons I wanted a base class is that I did not want to refactor. Refactoring using a base class is actually really simple.
Rename the original class to original_base.
Inherit from original_template inherits from original_base. Make sure to copy the contructor and pass through all the arguments to the base class.
The statement using original = original_template<your default case here> ensures that no other source code has to be modified just yet.
Applied to the example above I ended up doing something like this:
//Target compiled to Provider.lib
//Mixer.hpp
#pragma once
#include <iostream>
using namespace std;
struct normal { static void log() { cout << "normal\n"; } };
class Mixer_Base
{
private:
int mixcount;
public:
Mixer_Base(int count);
void noTemplateInvolved();
};
template<typename output = normal>
class Mixer_tempalte : public Mixer_Base
{
public:
Mixer_tempalte(int count) : Mixer_Base(count)
{}
void callingTemplate();
};
template<typename output>
void Mixer_tempalte<output>::callingTemplate()
{
output::log();
}
using Mixer = Mixer_tempalte<>;
//Mixer.cpp
#include "Mixer.hpp"
void Mixer_Base::noTemplateInvolved()
{
cout << "noTemplateInvolved\n";
}
Mixer_Base::Mixer_Base(int count) : mixcount(count)
{}
//Target Compiled to User.exe
//This target imports Provider.lib
#include <Provider\Mixer.hpp>
#pragma comment(lib, "Provider.lib")
struct awsome { static void log() { cout << "awsome\n"; } };
int main()
{
Mixer n(4);
n.callingTemplate();
n.noTemplateInvolved();
Mixer_tempalte<awsome> a(3);
a.callingTemplate();
a.noTemplateInvolved();
return 0;
}
In a way an annotation feels just like the base class feels. Everything in the base class is now annotated the way I wanted it to be, though this does not reduce inheritance.

How to make a function only seen by one function in c++?

How could I make a function only seen by the function that calls it?
define the function I want to hide as private function is not enough, as it could still be seen by other public functions in the class.
Now I use lambda expression to define anonymous function inside function. Is there any better solution?
Aside from using a lambda (which you've rejected), you could implement your function in its own compilation unit, and code the supporting function in an anonymous namespace within that compilation unit.
But that supporting function would be outside the class, so you'd have to pass it all the parameters it needed. That could become unwieldly though no worse than a long lambda capture list.
You can use a function object. For example(you can compile this, even in C++03):
#include <iostream> // only for output
class foo{
int bar(){return 0;} // Only foo can see this
public:
int operator()(){
return bar();
}
};
class baz{
public:
foo do_foo;
};
int main(){
baz a;
std::cout << a.do_foo() << std::endl;
}
the method bar is only visible by a foo.
P.S.: If you need foo to access members of baz, make it a friend.
A simmilar approach to cassiorenan would be to use static class functions and friends.
Something like this:
void Boss();
class Worker {
static void Test(){ return;}
friend void Boss();
};
void Boss(){
Worker::Test();
}
Though why would you want to do this, I don't know.
It is possible to define function inside a function without lambdas. Just define a struct that contains required function. This approach is not much better than using lambda, but at least this is straightforward and works with older compilers too.
int func() {
struct {
int hiddenFunc() {
return 1;
}
} h;
int a = h.hiddenFunc() + h.hiddenFunc();
return a;
}
As a slight variation from cassiorenan's solution, you could use a class containing one public static function (the visible function) and one static private function that could only be called from there. To avoid creation of objects of that class, it is enough to put a private constructor.
EDIT:
Per cassiorenan's comment, I can see that OP really needs methods and not functions. In that case, I would still use a dedicated class in a anonymous namespace to ensure it is not visible from elsewhere (even if my example is single file ...) friend to the class really used. So in below example, bar is the business class that would have a method with an externally hidden implementation (here relay_method), and foo is dedicated to the hidden method called with a pointer to the real object. In real world, the whole anonymous namespace and the implementation of the hidden method should be in the implementation file bar.cpp.
That way, the real implementation function priv_func can only be called from a bar object through bar::relay_method() and foo::bar_func(bar &).
#include <iostream>
class bar;
namespace {
class foo {
private:
static int priv_func(int i) {
return i * i;
}
foo() {}
public:
// only useful if true functions were needed
/* static int pub_func(int i, int j) {
return priv_func(i) + priv_func(j);
}*/
static void bar_func(bar& b);
};
}
class bar {
int x;
int x2;
public:
bar(int i): x(i) {}
void relay_method() {
foo::bar_func(*this);
}
friend class foo;
int getX2() const {
return x2;
}
};
void foo::bar_func(bar& b) {
b.x2 = foo::priv_func(b.x);
}
using namespace std;
int main() {
/* int i = foo::pub_func(3,4);
cout << i << endl;
// foo::priv_func(2); error access to private member of class foo
// foo f; */
bar b(2);
b.relay_method();
cout << b.getX2() << endl;
return 0;
}

C++ what to code if i put a class after main() function

I'm watching some video tutorials on C++ and i know you must define a function / class before it is used or called. But I like having my main() function at the top, and everything else below the main function. I know if i define a function below the main function I must declare it before it is used, but what about a class? What do I need to put above my main function to use my class below the main function.
#include <iostream>
using namespace std;
int main()
{
ClassOne one;
one.coolSaying();
return 0;
}
class ClassOne
{
public:
void coolSaying()
{
cout << "Cool stuff yo!" << endl;
}
};
I tried defining my class by placing this right before main():
class ClassOne;
but it doesn't work.
This is why header files are normally used in C++. When you're saying ClassOne one, the compiler needs to know what the class looks like to create an object of that type. It's not enough to know that the class exists somewhere (that is enough if all you want is a pointer). So the compiler needs to already have read the definition of the class.
Your class has to be defined before it is first used. Without putting it explicitly before main, the usual way is to create a header file. So you create ClassOne.h with the class declaration, and you have #include "ClassOne.h at the top of your file. In this situation the actual methods of the class would normally be in another source file, ClassOne.cpp.
A class MUST be "complete" when you create an instance of it. So there is no way you can use the class before you have defined the whole content of the class.
It is possible to do something like this:
class ClassOne;
ClassOne* make_class_one();
void use_class(ClassOne *x);
int main()
{
ClassOne* one = make_class_one();
use_class(one);
return 0;
}
class ClassOne
{
public:
void coolSaying()
{
cout << "Cool stuff yo!" << endl;
}
};
ClassOne* make_class_one()
{
return new ClassOne; // Bad idea, use uniqe_ptr, but I'm lazy.
}
void use_class(ClassOne *x)
{
x->coolSaying();
}
But in general, we don't want to do that.
One scenario where the class definition after the main() function makes sense:
#include <iostream>
using namespace std;
void f();
int main()
{
f();
return 0;
}
class ClassOne
{
public:
void coolSaying()
{
cout << "Cool stuff yo!" << endl;
}
};
void f()
{
ClassOne one;
one.coolSaying();
}
(note: all other answers are correct, but you may find this useful)
I discovered this idiom to invert the order of main and secondary function classes. I use to share small code with colleagues, everybody expects the core of the code (i.e. main) to be on top so they can edit it quickly. It works with classes and functions (without need of declaration) of course. Usually I can leave the preamble (first #includes) because those have include guards in most cases.
#include <iostream>
using namespace std;
#ifdef please_see_definitions_below_main
int main()
{
ClassOne one;
one.coolSaying();
return 0;
}
#else
class ClassOne
{
public:
void coolSaying()
{
cout << "Cool stuff yo!" << endl;
}
};
#define please_see_definitions_below_main
#include __FILE__
#endif
I use the tag please_see_definitions_below_main so it serves as comment also, but if you don't like it you can use something shorter, like AFTER.
You cannot create an actual instance of the type (variable, value member) until the type is fully defined, as its size is not known. There is no way around that, but there is a lot you can already do with a pointer to an incomplete type.

C++ inheritance problem with namespaces

OK, I have been looking about but can not for the wits of me find a reason to why this should not work:
Base class (misc/interface/handler.h)
#ifndef __t__MISC_VIRTUAL_HANDLER_H
#define __t__MISC_VIRTUAL_HANDLER_H
#pragma message("Starting with 'handler.h'")
namespace t {
namespace misc {
namespace interface {
class Handler {
public:
Handler();
virtual ~Handler();
virtual int setup() = 0;
virtual int teardown() = 0;
virtual int update() = 0;
protected:
private:
};
}
}
}
#pragma message("Ending with 'handler.h'")
#endif // __t__MISC_VIRTUAL_HANDLER_H
Derived class (graphics/handler.h):
#ifndef __t_GRAPHICS_HANDLER_H
#define __t_GRAPHICS_HANDLER_H
#include "../misc/interface/handler.h"
namespace t {
namespace graphics {
class Handler: public t::misc::interface::Handler {
public:
Handler();
virtual ~Handler();
int getResolutionX() { return m_resolutionX; }
int getResolutionY() { return m_resolutionY; }
bool getFullscreen() { return m_isFullscreen; }
protected:
private:
unsigned int m_resolutionX, m_resolutionY;
bool m_isFullscreen;
}; // class Handler
} // namespace graphics
} // namespace t
#endif // __t_GRAPHICS_HANDLER_H
... which seems rather trivial.
Derived class implementation (graphics/handler.cpp):
#include "handler.h"
t::graphics::Handler::Handler(): t::misc::interface::Handler() {
}
t::graphics::Handler::~Handler() {
}
... which too is should be really trivial, but yields the error:
src\graphics\handler.cpp|5|undefined reference to `t::misc::interface::Handler::Handler()'
I'm using MinGW with Code Blocks and what ever standard settings CB uses, I've tried building the same situation with test classes and that works as intended, both in same environment and Linux with vanilla g++.
I can't see any implementation of t::misc::interface::Handler::Handler() in your code - and it is going to be called by the inheriting class's constructor, so it needs an implementation. The linker can't find it, so it complains.
Just change:
Handler();
virtual ~Handler();
in the abstract class to:
Handler() {}
virtual ~Handler() {}
and you're ready to go.
As an aside, identifiers starting with two underscores are illegal in C++ (since they are reserved for the compiler). In practice, they shouldn’t be a problem in preprocessor but it’s best to err on the safe side here: simply don’t use them.