don't quite get the ampersand '&'

I don't quite get the ampersand '&' as in what is its function in the scanf/printf functions.

In the below code shows the prinf function printing out the char variable c without the ampersand.

Code:

# include <stdio.h>
void main(void)
{
    char c='c';
    printf("The character is %c", c); // this prints 'c'
    return;
}

The next code is the same as the above code but instead has the ampersand in the printf function.

Code:

# include <stdio.h>
void main(void)
{
    char c='c';
    printf("The character is %c", &c); // this prints an upside down triangle
    return;
}

Another code shown below stores and prints a string but without the ampersand in both the printf and the scanf.

Code:

# include <stdio.h>
void main(void)
{
    char string[10];
    scanf("%s", string); // Let's say I input 'Hello!'
    printf("Message: %s", string); // this prints 'Hello!'
    return;
}

This code has the ampersand in the scanf.

Code:

# include <stdio.h>
void main(void)
{
    char string[10];
    scanf("%s", &string); // Let's say I input 'Hello!'
    printf("Message: %s", string); // this prints 'Hello!'
    return;
}

When printing strings, whether the ampersand is present or not doesn't make a difference. And when dealing with char and int variables, the presence of the ampersand makes a difference. So I don't really get the ampersand '&' and would gladly appreciate it if someone could explain it to me

printf() needs a value to print, scanf() needs an address where to put the variable. scanf() needs to communicate a variety of data types back to the caller, best way to do that is by assigning a value to an address that was passed in by the caller and let it worry about the rest.

printf() doesn't need an ampersand anywhere in it, unless you request an address of a variable, to be printed. %p

scanf() will always need an ampersand, unless the variable already is the address of the variable it is about to change. Change being the key word here - you need that address to make the change.

I think your bottom line confusion is on arrays and pointers and how the compiler treats the array as a pointer to its first element when passed as a parameter. I also have a feeling you don't really understand what & means ( address-of). For more information you should read this:

Cprogramming.com FAQ > Pointers And Arrays (intermediate)

The ampersand in front of a variable represents the memory address where the variable is stored. So printf("%c", &x) prints the ASCII character assigned to the value of the memory address where x is stored. Since &x is probably "0xb880e8" or something, that printf statement is going to print something funky.

Compare these

Code:

char string[] = "hello world\n";
printf("The string=%s", string );
printf("The string=%s", &string[0] );

Now increment 0 to say 1, and see what happens...