CODEWITHSUNDEEP
c++functionpointersargumentsdefault
user765269
user765269

Reputation: 423

c++ function taking function pointer with a default function

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

Upvotes: 1

Views: 212

Answers (2)

aaronman
aaronman

Reputation: 18750

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.

Upvotes: 2

mark
mark

Reputation: 5459

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

Upvotes: 4

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