The Address of an Array

The name of the array is interpreted ad the address of the first element of an array,whereas applying the address operator yields the address of the whole array.

#include<iostream>
using namespace std;

int main(void)
{
	short tell[10];    //tell a array of 20 bytes
	cout<<sizeof(tell)<<endl; // the result of output is 20
	cout<<tell<<endl;    //displays &tell[0]
	cout<<&tell<<endl;    //displays address of whole array
	return 0;  
}

Numberically,these two addresses are the same,but conceptually &tell[0],and hence tell,is the address of a 2-byte block of memory,whereas &tell is the address of a 20-byte block of memory.
So the expression tell+1 adds 2 to the address value,whereas &tell+1 adds 20 to the address value.
tell is type pointer-to-short,or short ,and &tell is type pointer-to-array of 20-shorts or short ()[20].

short (*pas)[20] = &tell;    //pas pointer to array of 20 shorts

If you omit the parentheses,precedence rules would first associate [20] with pas,making pas an array of 20 pointers-to-short,so the parentheses are necessary.Next,if you wish to describe the type of a variable,you can use the declaration of that variable ad a guide and remove the variable name.Thus,the type of pas is short (*)[20].Also note that because pas is that set to &tell,*pas is equivalent to tell,so (*pas)[0] would be the first element of the tell array.