function pointers using '<' as an operator

Asked Dec 22 '14 at 16:47

Active Dec 22 '14 at 17:16

Viewed 83 times

I was learning function pointers and came across a piece of code here. The + is suppose to work as an operator/function

void FunctionA (float a, float b, float (*pt2Func)(float, float))
{
   float result = pt2Func(a, b);    // call using function pointer
}



void FunctionB ()
{
   Switch(2, 5, /* '+' specifies function 'Plus' to be executed */ '+');
}

So i experimented with it a little and this doesn't even compile.

Q. Why cant i use '<' as an operator here ?

Q. If not then does the user of the function need to write a separate code of compare two things whose comparision exists already...?

template <typename T>
bool compare (T a,T b,bool (*compareFunction) (T ,T ))
{
    return compareFunction (a,b);
}

int main ()
{
    int a=5,b=6;

    if (compare<int>(a,b,'<'))
        cout<<"Sucess";
    else
        cout<<"Post this in stack overflow ";

  return 0;
}

Update :

Some of the comments advice me to use this

(template <typename T, typename F> bool compare(T a, T b, F func))

But if i use this then this would only work for standard operators like std::less<int> . I want my function to work for both standard and non-standard operators (defined through their fucntions) both.

So the question still remains unanswered.

edited Jun 20 '20 at 09:12

Community

asked Dec 22 '14 at 16:47

Rohith R

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`'<'` isn't even a function. – user2864740 Dec 22 '14 at 16:49
so then is `'+'` one...? – Rohith R Dec 22 '14 at 16:49
`'<'` is of type `char`. It is not a function pointer. Neither are `'+'` or `'-'` or anything of a similar form. – user123 Dec 22 '14 at 16:50
then how come `'+'` is one...? – Rohith R Dec 22 '14 at 16:50
@PRP `'+'` is not a function either. – user2864740 Dec 22 '14 at 16:50
so u are saying the first piece of code also doesn't work.. – Rohith R Dec 22 '14 at 16:51
That is correct. Neither will work; enable/read warnings. – user2864740 Dec 22 '14 at 16:51
@PRP Yes, we´re saying that. It´s easy to write a real function to add two floats and use it there, but `+` alone just won´t work. – deviantfan Dec 22 '14 at 16:51
then what about the second part of my question...? – Rohith R Dec 22 '14 at 16:52
1

Make your function fully generic instead of being a function pointer, since functors are not function pointers. (`template bool compare(T a, T b, F func)`) Then you can use for example the standard functors `std::less`, `std::plus`, `std::minus`, et al. – cdhowie Dec 22 '14 at 16:52
1

The code that you linked in has a `switch` to decide which function pointer to return. You're looking at the wrong line with the `+` symbol. // this uses a switch Switch(2, 5, /* '+' specifies function 'Plus' to be executed */ '+'); // this uses a function pointer. No '+' here. Switch_With_Function_Pointer(2, 5, /* pointer to function 'Minus' */ &Minus); – xxbbcc Dec 22 '14 at 16:53
Simply :you cannot overload operators for built-in types. – SwiftMango Dec 22 '14 at 16:55
@cdhowie but then will your way work for non-standard operators defined as separate fucntions as well...? – Rohith R Dec 22 '14 at 17:00
@PRP If you mean functions, yes, `F` will simply be deduced to be a function pointer. – cdhowie Dec 22 '14 at 17:12
@cdhowie pls compile all this into a good example and put is as a well formatted answer...this will also help other having same problem – Rohith R Dec 22 '14 at 17:14
@PRP I can't because the question has been closed as a duplicate. – cdhowie Dec 22 '14 at 17:32

function pointers using '<' as an operator

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