don't quite get the ampersand '&'

[CHanabi]

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.

# 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.

# 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.

# 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.

# 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

[msh]

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.

[Adak]

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.

[claudiu]

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)

[Babkockdood]

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.

[Salem]

Compare these

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...