c++ function taking function pointer with a default function

Question

assume I have the following function

int watchVar(const char* var, const char* descriptor,
            Color (*colorfunc)(const char* var) = yellowColorFunc)

with

Color yellowColorFunc(const void* var){
    return Color::yellow();
}

I want to overload watchVar to accept functions whos parameters are char, int, float, etc, but do not want to create a default color function for each type.

g++ gives this error:

xpcc::glcd::Color (*)(const char*)' has type 'xpcc::glcd::Color(const void*)

Is there another way besides declaring colorfunc to take a void pointer and forcing the caller to cast the argument later himself?

Thanks

Please define "won't compile". Some compilation error would help. — Kelm
– Kelm, Commented Aug 6, 2013 at 21:50
-1 (it's only 2 rep points anyway) for not providing the compiler error message. — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Aug 6, 2013 at 21:53
@David downvotes on questions have been free of charge for quite some time now. — jrok
– jrok, Commented Aug 6, 2013 at 21:56
error: default argument for parameter of type xpcc::glcd::Color (*)(const char*)' has type 'xpcc::glcd::Color(const void*) — user765269
– user765269, Commented Aug 6, 2013 at 21:57
@jrok: Is it not -2 rep for the person that gets the downvote? — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Aug 6, 2013 at 22:10

mark · Accepted Answer · 2013-08-06 21:51:43Z

4

the function pointer is declared const char * but the yellowColorFunc is declared const void *

answered Aug 6, 2013 at 21:51

mark

5,4992 gold badges24 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

user765269 Over a year ago

I did that on purpose, so I could reuse yellowColorFunc for int,float etc.

David Rodríguez - dribeas Over a year ago

@user765269: This is the classic XY problem. You are asking how to improve your solution to a problem of which everyone is unaware. Take the time to formulate the real problem that you need to solve.

aaronman · Accepted Answer · 2013-08-06 22:00:26Z

2

Your problem is that your declaring a function pointer taking a const char * but yellowColorFun takes a const void *. If c++11 is available you can use std::function like so:
auto colorFunc = std::function<int(const char *,const char *,std::function<Color(const char*)>)>();

You said in a comment that you wanted to use the function for int,float and others, what your should do in that situation is use a templated function, you really don't wanna use void*'s in c++ very often.

answered Aug 6, 2013 at 22:00

aaronman

18.8k8 gold badges65 silver badges81 bronze badges

4 Comments

user765269 Over a year ago

Problem with templates is that I need the type for casting later, because the variable is printed out. RTTI is not available, this code is supposed to be embedded c++. Is there a typeid equivalent at compile time?

aaronman Over a year ago

@user765269 why do you need RTTI ints and floats support most of the same functions

Mooing Duck Over a year ago

@user765269 The compile time equivalent of RTTI and typeid is templates.

user765269 Over a year ago

You are right, I was overthinking the template version, that should totally do it.

Collectives™ on Stack Overflow

c++ function taking function pointer with a default function

2 Answers 2

2 Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related