Does this line declare a function? C++

Question

Hi all, I was reading litb's question about SFINAE here and I was wondering exactly what his code is declaring. A simpler (without the templates) example is below:

int (&a())[2];

What exactly is that declaring? What is the role of the &? To add to my confusion, if I declare the following instead

int b()[2];

I get an error about declaring a function that returns an array, while the first line has no such error (therefore, one would think the first declaration is not a function). However, if I try to assign a

a = a;

I get an error saying I'm attempting to assign the function a... so now it is a function. What exactly is this thing?

Thanks in advance

Answer 1

There's these awesome programs called cdecl and c++decl. They're very helpful for figuring out complicated declarations, especially for the byzantine forms that C and C++ use for function pointers.

tyler@kusari ~ $ c++decl
Type `help' or `?' for help
c++decl> explain int (&a())[2]
declare a as function returning reference to array 2 of int
c++decl> explain int b()[2]
declare b as function returning array 2 of int

a returns a reference, b does not.

Answer 2

int (&a())[2];

It declares a symbol a that is a function that takes no arguments and returns a reference to a two-element array of integers.

 int b()[2];

This declares a symbol b that is a function that takes no arguments and returns a two-element array of integers... this is impossible by the design of the language.

It is relatively simple: get an operator precedence chart, start the symbol name (a) and start applying the operators as you see from their precedence. Write down after each operation applied.

Answer 3

For future reference, you may find this link helpful when you have a particularly difficult C/C++ declaration to decipher:

How To Read C Declarations

For completeness, I will repeat what others have said to directly answer your question.

int (&a())[2];

...declares a to be a zero-argument function which returns a reference to an integer array of size 2. (Read the basic rules on the link above to have a clear understanding of how I came up with that.)

int b()[2];

...declares b to be a zero-argument function which returns an integer array of size two.

Hope this helps.