How to read a complicated type?

Question

I know that using type alias and typedef make the code so readable, less error-prone and easy modify. However in this example I want to know this complicated type:

#include <iostream>
#include <typeinfo>

char ( * (* x() )[] )();

int main(){

    std::cout << typeid(x).name() << std::endl;
}

As the rule of thumb says that I read it from inside through outside but I find it too confusing.

Can someone help me how to read it?

The output of GCC:

FPA_PFcvEvE

Answer 1

char        - returning a char                         *
  (*        - of pointers to                    *---v  ^
    (*      - returning a pointer to    *---v   ^   v  ^
       x()  - a function `x`          --^   v   ^   v  ^
    )[]     - an array                      *---^   v  ^
  )();      - functions                             *--^

And see https://itanium-cxx-abi.github.io/cxx-abi/abi.html#mangle.array-type , like:

FPA_PFcvEvE
^           - function (
 ^          - returning pointer
  ^         - to an array [
  ><        - with no dimension expression
   ^        - separator, end of array ]
    ^       - of pointers
     ^      - to functions (
      ^     - returning char
       ^    - (void) functions that take no arguments
        ^   - separator, end of function )
         ^  - (void) it's a function that takes on arguments
          ^ - separator, end of function )

Example:

#include <iostream>

char return_a() { return 'a'; }
char return_b() { return 'b'; }
char return_c() { return 'c'; }

char (* array_of_funcs_that_return_char[3])() = {
    return_a,
    return_b,
    return_c,
};

char (* (*pointer_to_array_of_funcs_that_return_char)[3])() = &array_of_funcs_that_return_char;

char ( * (* x() )[3] )() {
    return pointer_to_array_of_funcs_that_return_char;
}

int main() {
    std::cout << (*x())[1](); // outputs 'b'
}