What do the square brackets at the end of this code line mean?

Question

I'm fairly new to C++ and while coming across a reference site, I came across the following code snippet and I've been cracking my head to break it down ever since.

This is what I'm still trying to figure out:

int (*(*callbacks[5])(void))[3]

I've read through part of my C++ help books and (generally) understand the precedence flow. But seeing many operators bunched together has thrown me off and I'm rather confused. I've seen other examples and explanations (here), but the extra [3] subscript on the right just complicates things for me.

I was wondering on how I should approach such kinds of complex code, ie. where do I start, what order should I follow, etc.

Would really appreciate the help! Thanks!!

fwiw I've been writing in C++ for a long time, and if I saw that I'd get up and go get another coffee before trying to figure that one out — Cory Kramer, May 26 '17 at 18:14
You can try [this](https://cdecl.org/?q=int+%28*%28*callbacks%5B5%5D%29%28void%29%29%5B3%5D) — J. Calleja, May 26 '17 at 18:17

Miles Budnek · Answer 1 · 2017-05-26T18:31:48.230

10

That is an array of 5 pointers to functions that take no parameters and return a pointer to an array of 3 ints.

I only figured that out because cdecl.org told me. This is one of those places that a typedef would make things much easier to read:

typedef int (*PointerToArrayOfInts)[3];
typedef PointerToArrayOfInts (*FunctionReturningPointerToArray)(void);
FunctionReturningPointerToArray callbacks[5];

edited May 26 '17 at 18:31

answered May 26 '17 at 18:18

Miles Budnek

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No usage of the term "dereference"? ;) – Joe Phillips May 26 '17 at 18:29

score 9 · Answer 2 · answered May 26 '17 at 18:38

Here are the beginnings of a primitive introspective "pretty printer" which lets the C++ compiler itself break things down for you:

#include <iostream>

template <typename T>
struct introspect;

template <>
struct introspect<int> {
    static std::ostream& prettyPrint(std::ostream& os) { return os << "int"; }
};

template <typename T>
struct introspect<T*> {
    static std::ostream& prettyPrint(std::ostream& os) {
        os << "pointer to ";
        return introspect<T>::prettyPrint(os);
    }
};

template <typename T, std::size_t N>
struct introspect<T[N]> {
    static std::ostream& prettyPrint(std::ostream& os) {
        os << "array of " << N << " (";
        introspect<T>::prettyPrint(os);
        return os << ")";
    }
};

template <typename Res>
struct introspect<Res(void)> {
    static std::ostream& prettyPrint(std::ostream& os ) {
        os << "function returning (";
        introspect<Res>::prettyPrint(os);
        return os << ")";
    }
};

int main() {
    int (*(*callbacks[5])(void))[3];
    introspect<decltype(callbacks)>::prettyPrint(std::cout) << '\n';
    return 0;
}

Output:

array of 5 (pointer to function returning (pointer to array of 3 (int)))

This is an awesome idea, why'd no one thought of this before? — Passer By, May 26 '17 at 18:45
It's also easy to extend it with "reference to", "const", "volatile"; and with a bit of extra work, it also shouldn't be too hard to extend to print argument types of general function types. I just didn't happen to need those in this particular case. — Daniel Schepler, May 26 '17 at 19:00

score 1 · Answer 3 · edited May 26 '17 at 18:20

1

int (*(*callbacks[5])(void))[3]

declaring callbacks as array 5 of pointer to function (void) returning pointer to array 3 of int

edited May 26 '17 at 18:20

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answered May 26 '17 at 18:19

Shiva

17
3

What do the square brackets at the end of this code line mean?

3 Answers3

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