[malhajri2006]

What does "x" equals in each statement? and which statement increases the address and which one increases the value of x?

        #include<iostream>
        using namespace std;
         int main()
         {

        int p =9;
	int *x =&p;[/b]

        //(1)
	cout<<(int)x<<endl;
	cout<<(int)*x<<endl;

        //(2)
	*x++;
	cout<<(int)x<<endl;
	cout<<(int)*x<<endl;

        //(3)
	(*x)++;
	cout<<(int)x<<endl;
	cout<<(int)*x<<endl;

        //(4)
	*(x++);
	cout<<(int)x<<endl;
	cout<<(int)*x<<endl;

        //(5)
	*(x)++;
	cout<<(int)x<<endl;
	cout<<(int)*x<<endl;

	system("pause");
	return 0;

This post has been edited by malhajri2006: 07 April 2010 - 08:11 PM

[PlasticineGuy]

Which IDE do you use?

[malhajri2006]

I don't know what do u mean by IDE but I am using VC++ 2010

[PlasticineGuy]

Well then that's easy. Use the debugger. Press F10 and a yellow arrow will appear to the left of the code box. That represents the next statement to be executed. Go to the "Watch" window and add 2 entries: x and *x. They show the values as your code runs.

By the way, it's not possible to know the value of x, ever, because the OS allocates p to a new random spot in memory each time your program runs.

EDIT: Why don't you just put an end bracket at the end then run the code and find out? It tells you anyway.

This post has been edited by PlasticineGuy: 07 April 2010 - 08:23 PM

[malhajri2006]

Thank u that was helpful.

[kingfeanor]

PlasticineGuy, on 07 April 2010 - 07:18 PM, said:

By the way, it's not possible to know the value of x, ever, because the OS allocates p to a new random spot in memory each time your program runs.

That is only correct if his OS supports Address Space Layout Randomization. Not all OSes do. If he is running Windows XP or Server 2003 then this won't be happening. If he is running newer then it is possible but I believe it is a link option that must be enabled.