pointer problem

[baphomet90]

i have a program here. it is currently working the only problem is it gives the wrong answer


#include <cstdlib>
#include <iostream>
#include <stdio.h>

using namespace std;
int main()
{

int intarr[4] = {1,2,3,4 };
int i;
int *ptr;
ptr=intarr;

for(i=0; i<4; i++)
{
printf("element %d value = %d \n", i, *ptr);
ptr++;
cout<<ptr<<endl;

}


int adress;
printf("Please Enter an Address");
scanf("%X", &adress);

ptr = (int*)adress;

printf("The Value of %X is %d", adress, *ptr);


system("PAUSE");
return EXIT_SUCCESS;
}

The address that the program shows me is wrong how do i fix this

i used this
http://cplusplus.com/forum/beginner/1777/
as my source

[gcampton]

#include<stdio.h>
int main()
{
   int* ptr=(int*)0x22FF74;
   printf("%d",*ptr);

   return 0;
}

was the code that printed the value in the address, but to print the address of a variable(in this case a pointer) you need to print out the reference (aka the "address of") operator &

#include<stdio.h>
int main()
{
   int* ptr=(int*)0x22FF74;
   printf("%d\n",*ptr);

   printf("0x%X\n", &ptr);

   return 0;
}

so just change: printf("The Value of %X is %d", adress, *ptr);
to this: printf("The Value of %X is %d", &adress, *ptr);

and to nitpick, address is spelled with 2 d's, and instead of using system calls you should try and find better standard workarounds, such as cin.ignore(1000,'\n');

EXIT_SUCCESS could be better replaced with a 0; less typing anyway.

[baphomet90]

here is the output of my program:

element 0 value = 1
0x22ff64
element 1 value = 2
0x22ff68
element of 2 value = 3 0x22ff6c
element 3 value = 4
0x22ff70

Please Enter an Address 0x22ff6c
The Value of 22ff54 is4

I am expecting my program to answer 3


and when i asked 0x22ff64 my program should answer 1

currently my computer just answer random values

[cnoeval]

"it is currently working the only problem is it gives the wrong answer"

Ya gotta love it!
Only in Computer Science and Politics can you get away with a phrase like that.

[gcampton]

can you show the revised code, and please use code tags.