Hard time understanding precedence principle of char (*p)[sizeof(c)];

Question

I'm learning pointers and to challenge myself I tried dereferencing a pointer to a character array. Eventually this worked:

char (*p)[sizeof(c)];

Where c is an array c[]="something"

I'm having a hard time understanding how (*p)[sizeof(c)]; differs from *p[sizeof(c)];

Based on what I know presently (which is not much!) the computer is saying this in the case of (*p)[sizeof(c)];:

"p points to c! oh, and btw, p is an array of whatever sizeof(c) ends up being".

But even that seems strange to me, so I guess I am confused as to what is being constructed when the parenthesis are added.

Can someone explain?

Full code in context here:

#include <iostream>

using namespace std;

int main(int argc, char *argv[]) {

    char c[] = "something";

    char (*p)[sizeof(c)]; // this works
    // char *p[sizeof(c)]; // this doesn't?

    p = &c;

    cout << p << endl;  

    cout << *p << endl;
}

Answer 1

Types in C can be read with what is informally known as the right-left rule. You start from the name of the variable being declared, then go right while you can, then go left while you can, and start again. Parentheses stop you until all of their content has been considered.

In your example:

char (*p)[sizeof(c)];
       ^               p...                          (hit ')', turn around)
      ^                is a pointer...               (hit '(', turn around and remove '()')
         ^^^^^^^^^^^   to an array of `sizeof(c)`... (end of line, turn around)
^^^^                   chars.                        nothing left, we're done!

Without the parentheses, this becomes:

char *p[sizeof(c)];
      ^^^^^^^^^^^^     p is an array of `sizeof(c)`... (end of line, turn around)
^^^^^^                 pointers to char.

Which indeed is quite different.

Answer 2

In this definition

char (*p)[sizeof(c)];

you define a variable p which is a pointer (the contenct of the brackets are evaluated first) to an array of characters with size [sizeof c]. At this point in your code p does not point anywhere specific as it is not initialized yet.

With the other definition

char *p[sizeof(c)];

you define a variable p which is an array (the * is evaluated later) of pointers to character.

you have a pointer to an array vs an array of pointers.

Answer 3

Type mismatch in your assignment

p = &c;

is the problem.

c is an array of 10 chars and and &c is a pointer to an array of 10 chars.

In char (*p)[sizeof(c)];, p is a pointer to an array of sizeof(c) (10) chars. So, this matches with the type of &c.

In char *p[sizeof(c)];, p is an array of sizeof(c) (10) pointers. But this doesn't match with the type of &c.