Function Pointer

In my Apple interview, I was shown this syntax in C++ and was a little rusty with that.

https://stackoverflow.com/questions/2298242/callback-functions-in-c

// Define callback signature.
typedef void (*DoneCallback) (int reason, char *explanation);
 
// A method that takes a callback as argument.
void doSomeWorkWithCallback(DoneCallback done)
{
    ...
    if (done) {
       done(1, "Finished");
    }   
}
 
//////
 
// A callback
void myCallback(int reason, char *explanation)
{
    printf("Callback called with reason %d: %s", reason, explanation);
}
 
/////
 
// Put them together
doSomeWortkWithCallback(myCallback);

• https://stackoverflow.com/questions/5645375/how-do-i-make-a-function-asynchronous-in-c

Can you do something like this?

void doSomething(void *f) {
    f(5);
}
 
void g(int x){
    x+5;
}
 
doSomething(g);

NO

• void* is a generic pointer type in C++ that can point to any data type, but it does not retain information about the type of what it points to. Therefore, you cannot call f(5) since f is treated as a void*, and the compiler doesn’t know it’s meant to be a function pointer.

If you’re really stupid, you can do this (casting a void*)

• this is done a lot in embedded programming lol

 
void doSomething(void* f) { 
    // Cast void* to a function pointer type int(*)(int) 
    int (*funcPtr)(int) = (int (*)(int))f;
}

, to make it work you need

void doSomething(void (*f)(int)) {
    f(5);
}

Note that if g returned an integer, you’d need to do this

void doSomething(int (*f)(int)) {
    f(5);
}
 
int g(int x){
    return x+5;
}
 
doSomething(g);

Related

• Callback