can you tell me whats the output here

[yaraa]

int x=9;
int *yptr;
yptr=&x;
yptr+=2;

cout<<*yptr;
system("pause");
}

[vlad from moscow]

The code has undefined behavior. Nobody can say what will be outputed.:)

[gobiking]

It would be whatever value that is stored in the x's address + 2
you have int x = 9
and the pointer *yptr
&x is what obtains the address from x
yptr is obtaining this address as the value

so the in a quick example
yptr is = x's address + 2
so *yptr is now taking the value of this new address and will output it

example:(making up some fake address
yptr=100 + 2
*yptr=value(102)

[vlad from moscow]

@gobiking

It would be whatever value that is stored in the x's address + 2

More precisely it would be said

in the x's address + 2 * sizeof( int )

[yaraa]

Then it would be 11 is that right ??

[gobiking]

yeah i changed it right after i realized the yptr+=2 was not *yptr+=2 XD

[vlad from moscow]

As I said nobody can say what will be outputed. The code has undefined behavior. We do not know whether there is an object of type int at address &x + 2

[gobiking]

is this a homework assignment for you because you can just type in an ide and it will tell you what you will get.

[LB]

yaraa wrote:

Then it would be 11 is that right ??

No, it would be random memory.

Note that there is a very huge different between these two lines of code:

yptr += 2;  //change value in yptr
*yptr += 2; //change value at yptr