How is *x the same as x[]?

[winrawr]

I'm unsure of how declaring **x (a pointer to a pointer of x) is the same as declaring x[][] (a two-dimensional array of x's)... The theory I have is this:

int *x = {whatever}; (x is a pointer to the first int value)
*x (first array item, value pointed to by x)
x[2] (goes to the location pointed to by x, plus sizeof(int)*2, resulting in third item in the array)

yes?

--edit--

if my understanding is correct, is this what is referred to as "pointer arithmetic?"

[Intrade]

"Yes."

[VernonDozier]

"Yes."

Ditto. Yes with quotes around the word "yes". The phrasing of this sentence is a bit loose and inaccurate.I'm unsure of how declaring **x (a pointer to a pointer of x) is the same as declaring x[][] (a two-dimensional array of x's)... The theory I have is this:

**x is a pointer to a pointer to an integer not "a pointer to a pointer of x", and you can have a two-dimensional array of integers, but you can't have a two-dimensional array of x's

x[][] and **x are definitely not the same, but when used to dereference addresses, can accomplish the same thing.

#include<iostream>
using namespace std;


int main ()
{
    int y[4];
    y[0] = 5;
    y[1] = 10;
    y[2] = 15;
    y[3] = 20;
    
    int *x = y;
    int *z = x + 2;
    cout << *z << endl;
    cin.get ();
    return 0;
}

Here's an example, though it is a one-dimensional example, not a two-dimensional example. The line in red is an example of pointer arithmetic. The program will output 15.

[winrawr]

okay I see,
and yeah I guess the phrasing was a bit loose... but with a char array,
you can define it char x[]="rawrawrawrawrawr"; or char *x="rawrawrawrawrawr"; and have it behave the same, no? Isn't there really no difference between pointers and arrays (since there isn't any bounds checking of arrays...you could just go p[5] even if p is just used as a pointer to some object, couldn't you?) it seems like the [0] is exactly the same as *, except the []s just tell the compiler to add an amount to the address of the pointed-to address in the machine instruction (mov eax,[ebp+whatever])... so what's the difference between **x and x[][]?

--edit--

aside from the fact that int x[5] would allocate the appropriate amount of space to the stack (or heap), where as declaring it int *x cannot have this functionality.

[NicAx64]

you can't do one of the following even , if you thinks that way.
That's the difference

char ** chararrayarray =
{
	"My first string " ,
	"My Second string " ,
	"My third string",
	"My fourth string "
};

the meaning of this is pointer to pointer to a char , but it can continue until it found the zero character. so this is wrong then.
or

char [][] chararrayarray =
{
	"My first string " ,
	"My Second string " ,
	"My third string",
	"My fourth string "
};

the meaning of this is array of ( char array)

but

char *chararrayarray[] =
{
	"My first string " ,
	"My Second string " ,
	"My third string",
	"My fourth string "
};

is correct ! That's the difference. The meaning of this is array of (char *) , so you can openly do this.

but you can do this.

char * chararrayarray1[] =
{
	"My first string " ,
	"My Second string " ,
	"My third string",
	"My fourth string "
};

char **  chararrayarray =chararrayarray1 ;

and try this program .

#include <iostream>
using namespace std ;

char * chararrayarray1[] =
{
	"My first string " ,
	"My Second string " ,
	"My third string",
	"My fourth string "
};

char ** chararrayarray = chararrayarray1 ;


int main(int argc, char* argv[])
{
	cout << *(chararrayarray) <<endl ;
	cout << *(chararrayarray+1) << endl ;
	cout << *(chararrayarray+2) << endl;
	return 0;
}