Thread: macros: can i test if a function was defined?

  1. #1
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    macros: can i test if a function was defined?

    can i test if a function was defined for i see what function to use?
    if i call a not defined function, i will get 2 compilers errors. so for resolve the problem, i must use macros.
    i'm trying do some code but without success
    Code:
    #ifdef clsPointer->MouseClick            
                     clsPointer->MouseClick();
            #else
                MouseClick();
            #endif
    these code isn't completed, but for show what i mean.
    if i define the clsPointer->MouseClick, the compiler must use the clsPointer->MouseClick, but if sin't, the compiler must use MouseClick.
    so what i miss on these code?
    Last edited by joaquim; 04-29-2018 at 09:01 AM.

  2. #2
    Programming Wraith GReaper's Avatar
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    No, you can't. Not in the way you want, anyway.

    You could have an array of function pointers that its element is NULL when a function doesn't exist, like a C++ map. But of course there's no way to generate it automagically, it'd have to be maintained manually by the user.

    For example:
    Code:
    void foo()
    {
        // ....
    }
    
    void bar()
    {
        // ....
    }
    
    typedef void (*funcPtr)();
    
    std::map<std::string, funcPtr> myFunctions;
    myFunctions["foo"] = foo;
    myFunctions["bar"] = bar;
    
    if (myFunctions.find("bar") != myFunctions.end()) {
        myFunctions["bar"]();
    }
    EDIT: Wait... now that I think about it, aren't virtual methods saved as functions pointers inside an object? If you could check whether a virtual method existed or not, that would be great, but I don't know.
    Last edited by GReaper; 04-29-2018 at 09:53 AM.
    Devoted my life to programming...

  3. #3
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    No, you can't do this with macros, because macros are evaluated before function definitions are evaluated.

    There are several facilities in c++ that allow choosing between a specialized behavior, and a general purpose behavior if the specialization does not exist. These work such that at the call site calls the most general function, but a specialized implementation is invoked based on the type of th object calling it.

    For example
    Code:
    class DefaultContext {};
    struct ClsContext { //I'm only using a struct here to keep the example short. 
        Cls * clsPtr;
    };
    
    void mouseClick(DefaultContext)
    {
       mouseClick();
    }
    
    void mouseClick(ClsContext context)
    {
       context.clsPrt->mouseClick();
    }
    
    int main()
    {
      auto context = someFactoryToGetContext();
      mouseClick(context); // calls the correct function depending on the context type.
    }
    Last edited by King Mir; 04-29-2018 at 11:00 AM.
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

  4. #4
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    i'm sorry, never get notification(i use hotmail) and the options are on
    i have 1 code that tell me if the function is on class, but if it's just prototype, i will get the same
    i can share the code:
    Code:
    template <class Type>class TypeHasToString
    {
        // This type won't compile if the second template parameter isn't of type T,
        // so I can put a function pointer type in the first parameter and the function
        // itself in the second thus checking that the function has a specific signature.
        template <typename T, T> struct TypeCheck;
    
    
        typedef char Yes;
        typedef long No;
    
    
        // A helper struct to hold the declaration of the function pointer.
        // Change it if the function signature changes.
        template <typename T> struct ToString
        {
            typedef void (T::*fptr)();
        };
    
    
        template <typename T> static Yes HasToString(TypeCheck< typename ToString<T>::fptr, &T::MouseClick >*);//can i change these for receive a function name?
        template <typename T> static No  HasToString(...);
    
    
    public:
        static bool const value = (sizeof(HasToString<Type>(0)) == sizeof(Yes));
    };
    i must convert it to a macro(for choose another function name), but the value must be called.
    if the function have a prototype and not defined, i will get '1'.
    if the function isn't on class(prototype and definition), i will get zero.
    from here, can anyone advice me?

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