How can I call a function using a function pointer?

Question

Suppose I have these three functions:

bool A();
bool B();
bool C();

How do I call one of these functions conditionally using a function poi

Answer 1

You declare a function pointer variable for the given signature of your functions like this:

bool (* fnptr)();

you can assign it one of your functions:

fnptr = A;

and you can call it:

bool result = fnptr();

You might consider using typedefs to define a type for every distinct function signature you need. This will make the code easier to read and to maintain. i.e. for the signature of functions returning bool with no arguments this could be:

typdef bool (* BoolFn)();

and then you can use like this to declare the function pointer variable for this type:

BoolFn fnptr;

Answer 2

Note that when you say:

bool (*a)();

you are declaring a of type "pointer to function returning bool and taking an unspecified number of parameters". Assuming bool is defined (maybe you're using C99 and have included stdbool.h, or it may be a typedef), this may or may not be what you want.

The problem here is that there is no way for the compiler to now check if a is assigned to a correct value. The same problem exists with your function declarations. A(), B(), and C() are all declared as functions "returning bool and taking an unspecified number of parameters".

To see the kind of problems that may have, let's write a program:

#include <stdio.h>

int test_zero(void)
{
    return 42;
}

static int test_one(char *data)
{
    return printf("%s\n", data);
}

int main(void)
{
    /* a is of type "pointer to function returning int
       and taking unspecified number of parameters */
    int (*a)();

    /* b is of type "pointer to function returning int
       and taking no parameters */
    int (*b)(void);

    /* This is OK */
    a = test_zero;
    printf("a: %d\n", a());

    a = test_one; /* OK, since compiler doesn't check the parameters */
    printf("a: %d\n", a()); /* oops, wrong number of args */

    /* This is OK too */
    b = test_zero;
    printf("b: %d\n", b());

    /* The compiler now does type checking, and sees that the
       assignment is wrong, so it can warn us */
    b = test_one;
    printf("b: %d\n", b()); /* Wrong again */

    return 0;
}

When I compile the above with gcc, it says:

warning: assignment from incompatible pointer type

for the line b = test_one;, which is good. There is no warning for the corresponding assignment to a.

So, you should declare your functions as:

bool A(void);
bool B(void);
bool C(void);

And then the variable to hold the function should be declared as:

bool (*choice)(void);

Answer 3

If you need help with complex definitions, like

double (*(*pf)())[3][4];

take a look at my right-left rule here.

Answer 4

bool (*fptr)();

int main(void)
{
    ...
    ...
    printf("Enter your choice");
    scanf("%d",&a);
    switch(a)
    {
        case 0:
               fptr = A;
               break;
        case 1:
               fptr = B;
               break;
        case 2:
               fptr = C;
               break;
        case 3:
                break;
    }
    (*fptr)();
    return 0;
}

Your choice is stored in a. Then accordingly, functions are assigned in the function pointer. Finally, depending on your choice, the very same function is called to return the desired result.

Answer 5

I think your question has already been answered more than adequately, but it might be useful to point out explicitly that given a function pointer

void (*pf)(int foo, int bar);

the two calls

pf(1, 0);
(*pf)(1, 0);

are exactly equivalent in every way by definition. The choice of which to use is up to you, although it's a good idea to be consistent. For a long time, I preferred (*pf)(1, 0) because it seemed to me that it better reflected the type of pf, however in the last few years I've switched to pf(1, 0).

Answer 6

Declare your function pointer like this:

bool (*f)();
f = A;
f();