c++ - 如何阅读复杂的类型？

Question

我知道使用类型别名和 typedef 可以使代码更具可读性、不易出错且易于修改。但是在这个例子中我想知道这个复杂的类型:

#include <iostream>
#include <typeinfo>

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

int main(){

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

根据经验，我从内到外阅读它，但我发现它太困惑了。

谁能帮我读一下？

GCC 的输出:

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                             *--^

并查看 https://itanium-cxx-abi.github.io/cxx-abi/abi.html#mangle.array-type ，比如:

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 )

例子:

#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'
}