Comparing integer value to array.

[omnificient]

Should the following code work:

#include <stdio.h>

int main()
{
  int users[3]= {1, 2, 3};
  int user[8];
  int count;
  puts("Enter usernumber: ");
  scanf(" %d", &user);
  for (count = 0; count < 4; count++)
      {
       if (user == users[count])
          {
           puts("User found!");
          }
       }

  getchar();
  getchar();
  return 0;
}

Thank You.

[tabstop]

No.






I suppose you want reasons? Fine:
1. You only ever use user as a scalar (one integer), not an array of eight integers.
2. You try to access users[3], which doesn't exist.

[Dino]

When you code

 scanf(" %d", &user);

you are passing the address of an address of an array of ints.

[omnificient]

I'm sorry. I started out with a program that compared strings, but then I changed my mind and thought I'd better try it out with integers first.

One way or the other

 scanf(" %d", &user);

is neccesary for the program to function correcty.

[Dino]

Then you'll have to rewrite scanf(), or, change the definition of variable user.

Todd

[omnificient]

What do you mean exactly?

[Dino]

scanf() is not designed to accept a pointer to a pointer. That's what you have coded.

[Dino]

If you want to refer to a particular element of the user array, you can code

scanf(" %d", &user[3]) ;

or some such index into the array.

[omnificient]

I don't see your point, as this is exactly the way the user is asked for input in the book I'm using now. It will only work with the Adress-of operator, somehow.

I do see that &user should actually refer to the memory-adress of user.

Rather odd this.

[Dino]

Well, perhaps the BOOK MADE AN ERROR?

Try this:

  int user ;

instead of int user[8];

[omnificient]

I see that. So let me show you what I've got now:

#include <stdio.h>

int main()
{
  int users[3] = {1, 2, 3};
  int input;
  int count;
  puts("Enter user number: ");
  scanf(" %d", &input);
  for (count = 0; count < 4; count++)
      {
       if (input == users[count])
          {
           puts("User found.");
          }
       }
  getchar();
  getchar();
  return 0;
}

This works. The confusing thing for me now is the & in

  scanf(" %d", &input);

What is it doing there?

[Dino]

The & means "address of". Reading it like text, the scanf() says:

Read data from stdin, into the address of variable input, put a blank in the first position and then format the rest of the data as an integer (because the user told me "%d")

[omnificient]

Well I still don't understand completely, but you have at least given me some indication of what it's for there, even though I'm not working with pointers.

Thanks :-)

[Dino]

Oh, but you ARE working with pointers!! Anytime you use an array, you are working with pointers.

[Elysia]

Arrays are pointers? Hmmm
Let's see it this way. If you give someone a copy of something (your homepage?) and they modify it, it won't affect your original, will it? No, I don't think so.
However, if you were to give them the address of that homepage, then they could change your original homepage.

That's the whole pointer ordeal. By using pointers, you pass the address and not a copy.
But remember that you must not take the address of an array. The address of an array will be evaluated as a pointer to pointer (wrong).
To take the address of something, you use the & operator.