How to define a variable that is a function pointer that can assign other functions in c?

Question

So if I was given a the following 2 functions:

void printhex(int x, char y, char z)
{
    printf("%x%x%x,x,y,z);
}
void printdecimal( int x, char y ,char z)
{
    printf("%i%i%i,x,y,z);
}

How would i define a variable, that is a function pointer, that can be assigned either of the 2 given functions?

Also within the main function how would i determine if any command line arguments were entered after the program name when the program was executed. If anything was entered, how would I be able to assign the function pointer to the function printhex, but otherwise would assign the function to printdecimal?

Answer 1

You could declare a variable similarly to:

void (*printSomething)(int, char, char);

You could assign it to one of those functions with:

printSomething = &printhex;

or

printSomething = &printdecimal;

The & is completely optional, so you could also just write printSomething = printhex; for example.

You also didn't ask, but you can call the function through the pointer with (for example):

(*printSomething)(1, 2, 3);

The * is also optional, so you could just write:

printSomething(1, 2, 3);

If you need help with command line arguments as well, that should be a separate question.

Answer 2

A1) This is how the function pointer will look like:

void (*myFuncPtr)(int, char, char);

You then assign a function to it like this:

myFuncPtr = &printhex;

And call it like this:

(*myFuncPtr )(1, 'a', 'b');

A2) The argc variable of the main() will tell if any command line arguments were entered other than the program name itself. If argc > 1 , then argc - 1 other strings were passed when calling the program. This is because argc also counts the program name as one of the parameters. So something like this should work for you:

if (1 < argc)
{
    myFuncPtr = &printhex;
}
else
{
    myFuncPtr = &printdecimal;
}

Answer 3

Use a typedef

void printhex(int x, char y, char z) {
      printf("%x%x%x",x,y,z);
}

void printdecimal( int x, char y ,char z) {
      printf("%i%i%i",x,y,z);
}     

typedef void (*printer)(int x, char y ,char z);

printer hex = &printhex;
printer dec = &printdecimal;

Answer 4

If I understand you correctly, it should be simple:

void (* ptr)(int, char, char);
ptr = printhex;

(*ptr)(a, b, c); //this will call the function

Answer 5

Such variable would be declared as

void (*fptr)(int, char, char);

then assigned a value:

fptr = printhex;

and used to invoke a function:

fptr(13, 'a', 'b');

Or you may declare a new name for the function type:

typedef void (*PrintFuncPtr)(int, char, char);

then use it for a variable declaration:

PrintFuncPtr  fptr;

Answer 6

To define a function pointer you would use the following syntax:

return-type (*pointer name)(parameters);

in this case, you would use: void (*fp)(int, char, char); for a function pointer named fp suitable for the two functions you've defined.

Assigning a function pointer can be done like any other pointer, except you use the name of the function. For example, fp = printhex; would do exactly what it says: assigning the address of printhex to the pointer fp . You don't need to use the & operator for printhex because the identifier is implicitly converted to pointer-to-function.