Is this a pointer to a function pointer?

LB (13399)
I know that void(*)() is a pointer to a function that returns nothing and takes no parameters, but what is void(**)()? Is it a pointer to a pointer to a function?

Also, could someone please explain to me what a signature VARIABLE is? Such as this:
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//?
cout << typeid(void()).name() << endl;
 
void __cdecl(void)
What is a function signature when it is used such as return_type(param_type, ...) ?
Last edited on
Cubbi (4774)
Yes,
void(*)() is a pointer to function returning void
void(**)() is a pointer to pointer to function returning void
void() is a function returning void
These are types (just like int*, int**, and int). Like with any type, you can use typeid().name() to describe it.
Last edited on
LB (13399)
Can I do this? I don't have a compiler to check.
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typedef void() vf; //maybe typedef void vf(); ?

vf Func (AnotherFunc);
Func(); //calls it

void Call(vf &f)
{
    f();
}

Or am I interpreting your "just like [...] int" too literally?
Last edited on
Cubbi (4774)
Almost: can't initialize functions like variables, the rest is okay.

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#include <iostream>

typedef void vf();

vf Func;

void Call(vf &f)
{
    f();
}

void Func() {
    std::cout << "sure\n";
}

int main()
{
    Func();
    Call(Func);
}


And yes, there are other limitations on function types: can't pass by value for example, which is why we have function objects.
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LB (13399)
Oh! So this whole time, function prototypes were actually...wow, I never knew. Thanks Cubbi!
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