array

[usama sadaqat]

#include "stdafx.h"
#include<iostream>
#include<conio.h>
#include<ctime>
using namespace std;
int *p[10][20];//
int _tmain(int argc, _TCHAR* argv[])
{
    cout<<sizeof(*p)<<endl;
    getch();
    return 0;


}

output: 80
but  i cannot understand how it work.

[phorce]

What don't you understand?

You are creating a pointer to an array and trying to output the size, this is not allowed sice the compiler does not know what the pointer is pointing to.

In your cose, you woud not not need a pointer, it would just work lik this:

#include<iostream>

using namespace std;
int p[10][20];//
int main()
{
    cout<<sizeof(p)/sizeof(int)<<endl;
    //getch();
    return 0;
} //output: 200

[usama sadaqat]

ok sir

[sepp2k]

You are creating a pointer to an array

No, he's not. He's creating an array of arrays of pointers.

and trying to output the size, this is not allowed sice the compiler does not know what the pointer is pointing to.

Yes, it's allowed and yes, the compiler knows perfectly well what p (or rather the pointer that p decays to) is pointing to: to a 20 element arrays of int pointers. So it returns sizeof(int*) * 20

And if he had created a pointer to an array (by writing int (*p)[10][20] instead of int *p[10][20]), it'd still be allowed and the compiler would still know what p points to: a 10x20 array of int. So in that case it would return sizeof(int) * 200.

[Ancient Dragon]

sizeof(int) * 200.

Thats the same as sizeof(*p)

[sepp2k]

I'm sorry, I do not understand your reply. I said that "in that case it [i.e. sizeof(*p)] would return sizeof(int) * 200". So yes, that's the same as sizeof(*p) (if p had been defined as a pointer to an array) - that's what I was saying.