what does this mean char (*(*a[4])())[5]?

Question

Hi I came across the question in \"Test your skills in c++\".

Please let me know what does it mean with an example?

Edited Section

Answer 1

I cheated by removing what I think is an extra right-parenthesis and pasting the result into cdecl.

declare a as array 4 of pointer to function returning pointer to array 5 of char

Answer 2

Following the spiral rule (as linked to by chris), and starting with the identifier:

a

...is...

a[4]

...an array of 4...

*a[4]

...pointers to...

(*a[4])()

...a function taking no parameters...

*(*a[4])()

...returning pointer to...

(*(*a[4])())[5]

...an array of five...

char (*(*a[4])())[5]

...chars.

Sidenote: Go give the architect who came up with this a good dressing-down, then find the programmer who wrote this code without a comment explaining it and give him a good dressing-down. In case this was given to you as a homework, tell your teacher that he should have instructed you on how to use cdecl instead, or how to design code in a way that it doesn't look like madman scrawlings, instead of wasting your time with this.

Answer 3

And another example... of what to never ever do in anything other than an example.

#include <iostream>

typedef char stuff[5];
stuff stuffarray[4] = { "This", "Is", "Bad", "Code" };

stuff* funcThis()   { return &(stuffarray[0]); }
stuff* funcIs()     { return &(stuffarray[1]); }
stuff* funcBad()    { return &(stuffarray[2]); }
stuff* funcCode()   { return &(stuffarray[3]); }

int main()
{
    char (*(*a[4])())[5] = { funcThis, funcIs, funcBad, funcCode };
    for(int i = 0; i < 4; ++i)
    {
        std::cout << *a[i]() << std::endl;
    }
    return 0;
}

Answer 4

And here's an example ...

#include <stdio.h>

char a[5] = "abcd"; 
char b[5] = "bcde"; 
char c[5] = "cdef"; 
char d[5] = "defg"; 

char (*f1())[5] { return &a; }
char (*f2())[5] { return &b; }
char (*f3())[5] { return &c; }
char (*f4())[5] { return &d; }

int main()
{
        char (*(*a[4])())[5] = { &f1, &f2, &f3, &f4 };
        for (int i = 0; i < 4; i++)
                printf("%s\n", *a[i]());
        return 0;
}