description of mixing pointer and array syntax

gistfile1.txt

[18:15:43]  <Chris>	 Norrin: it's, like all C declarations, "declaration follows use".

[18:16:14]  <Chris>	 So consider char (*x)[42]; You can then reason that *x is a char[42]. And (*x)[3] is a char.

[18:16:46]  <Chris>	 Norrin: and in the case of char *x[42]; you can say that x[3] is a char * and *x[3] is a char.

so [] has precedence over *.  hence why the () are required depending on intention. 

[18:14:32]  <Chris>	 Norrin: char *(*a)[42]; is a pointer to array of 42 char pointers.

[18:18:43]  <Chris>	 Norrin: you can also have a pointer to a variable length array though: int f(size_t s, int(*x)[s]);
[18:18:47]  <Norrin>	 Chris,  from anther var that tells you the size
[18:18:57]  <Chris>	 Norrin: okay, sure, so put a variable in, like the above.

[18:23:02]  <Chris>	 Norrin: in all C declarations, they follow the same pattern as using the objects.
[18:23:19]  <Chris>	 Norrin: int x; // x is an int.
[18:23:24]  <Chris>	 Norrin: int *x; // *x is an int.
[18:23:37]  <Chris>	 Norrin: int *x[42]; // x[n] is an int *
[18:23:46]  <Chris>	 Norrin: int (*x)[3]; // *x is an int[3]
[18:23:51]  <Chris>	 This is declaration follows use.
[19:54:14]  <Norrin>	 confusing that [ ] goes after identifier but * goes before it and after the target type


[20:32:05]  <Zhivago>     Just wait until you encounter something like int (*(*p)(int))[10];
[20:48:09]  <Sadale>	 int (*(*p)(int))[10] is a pointer to a function that receives and int and return a pointer to array of 10 of int?
[20:50:28]  <Norrin>	 pointer to a function that returns a pointer to an int[10]
[20:59:04]  <Norrin>	 how would you write p is a pointer to an array of pointers to int f(int) ?
[21:01:15]  <Norrin>	 int(int)(*p)[10]?
[21:01:33]  <Chris>	 int (*(*p)[10])(int)

[21:11:38]  <Norrin>	 Chris, if i hadn't forgot about it being an array of pointers to the function, i would have put int*(int)(*p)[10]
[21:12:01]  <Chris>	 that's a syntax error
[21:12:13]  <Chris>	 (and doesn't make sense)
[21:19:27]  <Chris>	 so how do you deference a function pointer?
[21:20:36]  <Chris>	 Actually (*f)(params)
[21:20:52]  <Chris>	 Because *f(params) is like *(f(params)) and will dereference the pointer the function returns.
[21:22:01]  <Chris>	 So now you can declare function pointers the same way.
[21:22:45]  <Chris>	 And this is why we declare a function returning an int * as int *f(something)
[21:22:51]  <Chris>	 Because *f(something) will be an int.