why the code can compile -- about function pointer

Hello everyone,


I am learning code from others, but I do not know why the following code section can compile?

Why assignment p = new (void (*[3])()) is ok? Are the left side and right side of assignment having compatible type?

Code:

---

void (**p)();

int main()
{
        p = new (void (*[3])());
        return 0;
}

---

thanks in advance,
George

p is defined as having type pointer-to-pointer-to-F, where F is function-returning-void-with-no-arguments.

void (*[3])() is an array of pointer-to-F; new will return a pointer to the first element, which is therefore of type pointer-to-pointer-to-F.

I suppose typedefs can help a lot to clear up things:

Code:

---

    typedef void(*func)();
    func* p = new func[3];

---

Thanks tabstop,


Your reply is clear.

The code,

Code:

---

        new void (*[3])();

---

will create an array of 3 function pointers, and the function pointer points to functions whose prototype is -- return value void and empty parameter list?

Quote:

---

Originally Posted by tabstop

p is defined as having type pointer-to-pointer-to-F, where F is function-returning-void-with-no-arguments.

void (*[3])() is an array of pointer-to-F; new will return a pointer to the first element, which is therefore of type pointer-to-pointer-to-F.

---

Thanks anon,


I agree. If written in that way, I will have no question. :-)

Quote:

---

Originally Posted by anon

I suppose typedefs can help a lot to clear up things:

Code:

---

    typedef void(*func)();
    func* p = new func[3];

---

---

have a good weekend,
George