C++: Pass function by reference

unihiekka · 11-12-2008, 03:14 AM

Is it possible to pass a function or a function handle by reference in c++?

I have the following situation: several functions, let's call them f(t,x), g(t,x) and h(t,x), where t and x are all variables of the same type and length. The only difference between f, g, h is that they give different function values. Now I want to pass f, g, h by reference so that another function can manipulate them without the need to rewrite the function for f, g, h:

void manip(fun,t,x)
{
/* do something with fun at t and x where fun can be f, g, h (or at least some function with the same parameters as f, g or h. */
}

Is that possible or am I thinking too much in terms of MATLAB now?

graemef · 11-12-2008, 03:35 AM

you can certainly do it, how you do it will depend if the function is a non static member function or not.

You first need to get the function type determined:
non-member function
int (*)(int,float)
member function
int (MyClass::*)(int,float)

unihiekka · 11-12-2008, 04:26 AM

They are called function pointers, quite logically... as if I could not think of that before.

unihiekka · 11-30-2008, 10:30 AM

OK, so if I have a simple (scalar) function that I pass to another function, I could use (CalcX is supposed to be a function that returns a double when given x):

Quote:

typedef double (*PTRFUN)(double);

double somefunction((*f)(double))
{
double value;
x = CalcX(x);
value = (*f)(x) + x;
return value;
}

So, I use "(*f)(x)" to use the value of f at x, which I have initialized in the main function with

Quote:

PTRFUN f;

. The function somefunction for f(x) is called by

Quote:

somefunction(f);

Defining an array of function pointers is easy:

Quote:

PTRFUN fun[n];

where n is some known integer value. I can then call

Quote:

somefunction(f[m]);

if I want to do that for the mth element of the function array.

But is it possible to send the whole array of functions f to somefunction? How would I then call somefunction for f[0] ... f[n-1], and how would I use f and its elements in the function itself: (*f[n])(x) or something more exotic?!

Further, is it possible to use a dynamic array of function pointers? That is, something like:

Quote:

PTRFUN* dynptrfun = NULL;
int n=/* some value which might be determined at runtime */;
dynptrfun = new PTRFUN[n];
/* initialize elements of dynamic array */
/* do something with it */
delete [] dynptrfun;
dynptrfun = NULL;

johnsfine · 11-30-2008, 10:48 AM

Quote:

Originally Posted by unihiekka

But is it possible to send the whole array of functions f to somefunction? How would I then call somefunction for f[0] ... f[n-1], and how would I use f and its elements in the function itself: (*f[n])(x)

That looks right to me.

Quote:

Further, is it possible to use a dynamic array of function pointers? That is, something like:

You seem to already understand it all. You could confirm you are correct more quickly by trying it than by asking a question here.

You also seem to understand the idea of using a typedef to help, but I'll stress that idea.

As you get into complicated types (such as an array of pointers for functions with a specific signature) C and C++ have general methods for cramming the whole mess into one type specification, but it is a bad idea to use them. Instead use multiple typedefs, each one adding just one layer to what the previous defined. That makes it much easier to understand the error messages when you get something wrong and it makes it much easier to read the code later and understand what you did.

ta0kira · 11-30-2008, 01:35 PM

Quote:

Originally Posted by unihiekka

They are called function pointers, quite logically... as if I could not think of that before.

Well, you can also pass them as references if you use the right type specification:

Code:

void function();
//...
void(&reference)() = function;
reference();

That isn't useful unless you won't need to change what it points to, and it obviously can't be used in an array.
ta0kira