Syntax for declaring a function that uses function pointers

[Sythion]

Hey all,

I'm new to c++ and function pointers. I think I have all my function pointers defined correctly, but I can't find out how to declare a function (constructor, actually) that uses these pointers as arguments.

What is the syntax for this?

in example.h

...
class example
{
	private:
		(const char**) (funcPtr1);
		(const char**) (funcPtr2);
	public:
		example(const char**, const char**); //this is wrong
}
...

in example.cpp

example::example(const char** (*funcPtr1)(), const char** (*funcPtr2)())
{
	const char** arr1= (*funcPtr1)();
	const char** arr2= (*funcPtr2)();
}

[zb0S216C]

Sythion wrote:

but I can't find out how to declare a function (constructor, actually) that uses these pointers as arguments. (sic)

Do you mean like this:

example( const char( **FunctionOne )( void ), const char( **FunctionTwo )( void ) )
{
    FunctionOne( );
    FunctionTwo( );
}

_Wazzak

[andywestken]

I would use

example::example(const char** (*funcPtr1)(), const char** (*funcPtr2)());

void parameters are not required for C++ code; they are supported to allow compilation of code by C as well as C++ compilers. I omit them from all C++ code.

Your class definition just needs to follow suit.

class example
{
	private:
		const char** (*m_funcPtr1)();
		const char** (*m_funcPtr2)();
	public:
		example(const char** (*funcPtr1)(), const char** (*funcPtr2)());
};

example::example(const char** (*funcPtr1)(), const char** (*funcPtr2)())
{
	const char** arr1= (*funcPtr1)();
	const char** arr2= (*funcPtr2)();
	
	m_funcPtr1 = funcPtr1;
	m_funcPtr1 = funcPtr2;
}

[andywestken]

PS A habit I have picked up from assorted people (authors, teams leads, ...) is to always use function pointers through typedefs. I find it makes things a lot clearer.

class example
{
	private:
		typedef const char** (*PFNEXAMPLE)();

		PFNEXAMPLE m_funcPtr1;
		PFNEXAMPLE m_funcPtr2;
	public:
		example(PFNEXAMPLE funcPtr1, PFNEXAMPLE funcPtr2);
};

example::example(PFNEXAMPLE funcPtr1, PFNEXAMPLE funcPtr2)
{
	const char** arr1= (*funcPtr1)();
	const char** arr2= (*funcPtr2)();
	
	m_funcPtr1 = funcPtr1;
	m_funcPtr1 = funcPtr2;
}

[Sythion]

Thanks!

I changed the code as follows, and it compiles (although I'm still a bit away from actually doing somehting useful with it).

in example.h

...
class example
{
	private:
		//(const char**) (funcPtr1);
		//(const char**) (funcPtr2);
	public:
		example(const char**(funcPtr1)(), const char**(funcPtr2)()); //this is wrong
}
...

in example.cpp

example::example(const char** (*funcPtr1)(), const char** (*funcPtr2)())
{
	const char** arr1= (*funcPtr1)();
	const char** arr2= (*funcPtr2)();
}

It's very odd that there is an identifier included in the declaration though. What's the purpose of that?

[zb0S216C]

andywestken wrote:

void parameters are not required for C++ code

True, but some people prefer it as a preference since it explicitly indicates that the function doesn't require arguments. The problem I see with type-defined function pointers is that I have to go to their definition and see it that function pointer requires arguments, which is annoying.

_Wazzak

[andywestken]

I am aware that some people like to use void, but I don't get why "void" makes it clearer (or more explicit) that a function doesn't take a parameter as opposed to it not taking a parameter.

I find using typedefs makes the use of function pointer far less painful, esp. with more complicated functions. In general, the typedef is close to the use. When it isn't, I use a comment. Part of the reason I don't share your concern is that I use function pointers very sparingly, and then only ever through an excapsulating class. So once the class is in use, you don't go near the function pointer again.