C++ function parameters wont compile

Question

So I have a function with the prototype

double Solution::bisect
(
    double xLeft,
    double xRight,
    double epsilon,
    double f(double x, EquCoeffs coeffsStruct),
    bool &error
);

where function f is prototyped like this

double Solution::f(double x, EquCoeffs coeffsStruct);

The error I am getting in Visual Studio is argument of type "double(Solution::*)(double x, Solution::EquCoeffs coeffsStruct)" is incompatible with parameter of type "double(*)(double x, Solution::EquCoeffs coeffsStruct)"

which occurs when I try to call bisect(xLeft, xRight, epsilon, f, error); where xRight xLeft epsilon, are type double and error is type bool.

Why won't this compile? I'm not sure I understand the error. How is a double(Solution::*) different than double(*)?

Please show a [mcve]. The error is from how you call the function. You try to call it with a pointer to a member function while it expects a pointer to a free function — 463035818_is_not_an_ai, Apr 05 '17 at 16:11
How can I make it expect a pointer to a member function instead? — Bassinator, Apr 05 '17 at 16:12
Okay, but when I change the signature of the function to expect `Solution::f`, I get `qualified name is not allowed`. I did some googling, and apparently I am not allowed to specify a class name here? What gives? — Bassinator, Apr 05 '17 at 16:20

463035818_is_not_an_ai · Accepted Answer · 2017-04-05T16:49:34.303

Function pointers can be tedious to work with, this is where typedefs come in handy:

typdef R (*FREE_FUNC)(ARGS);

makes FREE_FUNC an alias for function pointers of type R(*)(ARGS), ie functions with ARGS parameter and R return type. It may look a bit strange, that the new name appears in the middle on the typdef. However, everywhere else using function pointers is then much nicer, for example

typedef void (*LOGGER)(R,ARGS);
R decorate(FREE_FUNC func,LOGGER logger,ARGS args){
    R result = func(args);
    logger(result,args);
    return result;
}

Member function pointers are different, because you need an object to call them:

struct Foo {
    void foo(){std::cout << "moo \n";}
    void moo(){std::cout << "xyz \n";}
};

typedef void (Foo::*FOO_MEMBER_FUNC)();

void fooCaller(Foo f,FOO_MEMBER_FUNC mf){
    (f.*mf)();
    // ... or ...
    Foo* f_ptr = &f;
    (f_ptr->*mf)();
}

int main() {
    FOO_MEMBER_FUNC foo_ptr = &Foo::foo;
    FOO_MEMBER_FUNC moo_ptr = &Foo::moo;
    Foo f;
    fooCaller(f,foo_ptr);
    fooCaller(f,moo_ptr);
}

I should also mention, that there are better ways to pass functions. You can take a look at std::function.

I'm aware; unfortunately, my requirements dictate that I do it by passing a function parameter in such a manner. — Bassinator, Apr 05 '17 at 16:56
@Airhead not sure if I understand you. If the answer didnt help, you need to provide the code ([mcve]) — 463035818_is_not_an_ai, Apr 05 '17 at 16:58
My solution ended up being just to use a free function instead. Thanks though! — Bassinator, Apr 05 '17 at 16:59

C++ function parameters wont compile

1 Answers1