Address of (&) Operator and its use

Hello.
i'm a newbie and i'm learning c... i had a question in mind in i thought here's a good place to ask...

consider the following piece of code:

Code:

---

#include <stdio.h>
void main (void)
{
    char a[10];
    scanf("%s",a);
    printf("character: %s \n",a);

    int b;
    scanf("%d",&b);
    printf("int: %d \n",b);
}

---

my question is that, "logically" i can't understand why for scanf("%d",&b) "&b" has been used, and in scanf("%s",a), "a" without "&" operator has been used? although in printf("int: %d \n",b), b has been used without "&" operator...

what's the use of & operator? i would appreciate it if someone could answer this question "logically"...
i asked some of my colleges , and they told me that scanf() only accepts pointers, and when i use scanf() i should use & operator, i asked for reason, they replied that's the way it is... i can't find this answer logical... if that's the case? is it the same thing for printf(). if so, then why printf("int: %d \n",b); is not using "&" operator?

& is the address of operator. The "address of" a variable is the same as a pointer to the value of that variable. As you hopefully understand, a pointer's value is the memory address where another value (of a certain type) can be found:

Code:

---

int n = 5, *ptr = &n;

---

What has you confused here is the convention that the name of a string variable -- or, in fact, the name of any kind of array, which a C-string is a char array terminated with '\0' -- works as a pointer to the first element of the array.

Code:

---

char str[]="hey", *p = str;

---

Notice no & needed here (in fact, it would be incorrect). You could use "*p = &str[0]" (the address of the first element).

There is a very logical reason why scanf needs an address to an integer rather than the integer. It's because scanf reads the input and then *rerturns* the integer in the variable the pointer points to. If it wouldn't be a pointer, the called function wouldn't be able to write to the variable, as it could only modify a local copy. Imagine when you pass variables to your own functions and you change them in these functions: they change in the function, but not in the calling scope, because in the function copies rather than the actual variables are used.

i understand now :)
thanks