difference between a pointer and a pointer array

Question

I made a code

#include <iostream>
#include<conio.h>
using namespace std;

void main()
{
    int *x,*y;
    x=new int[1];
    y=new int;
    cin>>y;   //Gives error probably because y is a pointer and not a variable
    cin>>*y                 //works fine
    cin>>x[0]>>x[1];
    cout<<x[0]<<x[1];
    cout<<*x[0];         //gives error
    cout<<y;
    cout<<*y;

    getch();

}

gives error.why?I remember i declared x as a pointer array and now i m doing the same i did with *y.Does it mean that a pointer array becomes a variable?plz help!

x and y are both variables of type pointer to int. The reason for the first error is that istream has no overloaded extraction operator for reading a pointer (see cplusplus.com/reference/istream/istream/operator%3E%3E), and the reason for the second error is that you are attempting to dereference something not of pointer type (x[0] has type int). — warrenm
– warrenm, Commented Mar 21, 2013 at 4:57
@warrenm i thought i created array of pointers that are pointing to int and not array of int. what can i do to create array of pointers? — Smatik
– Smatik, Commented Mar 21, 2013 at 5:03
@Smatik, I'm just saying it should. Any return type other than int is deemed undefined behaviour by the C++ standard. — Qaz
– Qaz, Commented Mar 21, 2013 at 5:04
@chris void main is not giving any error. Why should it give? i have not specified any return. — Smatik
– Smatik, Commented Mar 21, 2013 at 5:05

Serdalis · Accepted Answer · 2013-03-21 05:01:24Z

1

What you are actually doing with that line of code is similar to:

cout<<**x;

Because using x[0] will dereference the 0th element of x.

As you can see by your definition of x, x is just a pointer, not a pointer to a pointer, so dereferencing it twice will not work since you are trying to dereference a variable.

What the line:

x=new int[1];

is actually doing is just saying "assign an array of ints, size 1 to this pointer", which will just make x point to a block of memory big enough to store 1 int.

edited Mar 21, 2013 at 5:01

answered Mar 21, 2013 at 4:55

Serdalis

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DavidA · Accepted Answer · 2013-03-21 04:54:59Z

1

x is a pointer to an int. You have allocated an array of ints, which is a single int long. Therefore x[0] is an int and *x is an int. However, *x[0] means you are saying that x[0] is a pointer which you are dereferencing. However, it isn't a pointer, it is an int. That is why there is an error.

answered Mar 21, 2013 at 4:54

DavidA

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SeasonalShot · Accepted Answer · 2013-03-21 04:58:42Z

1

The meaning of the array:

x[0]

is equivalent to *(x+0);

As you know array is array is nothing but pointer in its root.

So any array that has x[a] or x[a][b] can be expanded as

*(x+a) or *(*(x+a)+b)

Based on this , i hope you found your answer.

answered Mar 21, 2013 at 4:58

SeasonalShot

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1 Comment

Benjamin Lindley Over a year ago

An array is not a pointer, though it can be implicitly converted to one.

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