Alright, I think the title is sufficiently descriptive (yet confusing, sorry).
I'm reading this library: Timer1.
In the header file there is a public member pointer to a function as follows:
class TimerOne
{
public:
void (*isrCallback)(); // C-style ptr to `void(void)` function
};
There exists an instantiated object of the TimerOne class, called "Timer1".
Timer1 calls the function as follows:
Timer1.isrCallback();
How is this correct? I am familiar with calling functions via function pointers by using the dereference operator.
Ex:
(*myFunc)();
So I would have expected the above call via the object to be something more like:
(*Timer1.isrCallback)();
So, what are the acceptable options for calling functions via function pointers, as both stand-alone function pointers and members of an object?
These are all valid and fine ways to call a function pointer:
myFuncPtr();
(*myFuncPtr)();
(**myFuncPtr)();
(***myFuncPtr)();
// etc.
(**********************************f)(); // also valid
Things you can do with a function pointer.
1: The first is calling the function via explicit dereference:
int myfunc(int n)
{
}
int (*myfptr)(int) = myfunc;
(*myfptr)(nValue); // call function myfunc(nValue) through myfptr.
2: The second way is via implicit dereference:
int myfunc(int n)
{
}
int (*myfptr)(int) = myfunc;
myfptr(nValue); // call function myfunc(nValue) through myfptr.
As you can see, the implicit dereference method looks just like a normal function call -- which is what you’d expect, since function are simply implicitly convertible to function pointers!!
In your code:
void foo()
{
cout << "hi" << endl;
}
class TimerOne
{
public:
void(*isrCallback)();
};
int main()
{
TimerOne Timer1;
Timer1.isrCallback = &foo; //Assigning the address
//Timer1.isrCallback = foo; //We could use this statement as well, it simply proves function are simply implicitly convertible to function pointers. Just like arrays decay to pointer.
Timer1.isrCallback(); //Implicit dereference
(*Timer1.isrCallback)(); //Explicit dereference
return 0;
}
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