Why doesn't scanf need an ampersand for strings and also works fine in printf (in C)?

Question

I am learning about strings in C now.

How come to use scanf to get a string you can do

scanf("%s",str1);

and for printf you can do

printf("The string is %s\n", str1);

I understand that for scanf it is because the string is just a character array which is a pointer, but for printf, how is it that you can just put the variable name just like you would for an int or float?

Answer 1

Err, nm, I'm tired :(

Answer 2

scanf needs the address of the variable to read into, and string buffers are already represented as addresses (pointer to a location in memory, or an array that decomposes into a pointer).

printf does the same, treating %s as a pointer-to-string.

Answer 3

In C, variables that are arrays become a pointer to the first element of the array when used as function arguments -- so your scanf() sees a pointer to memory (assuming "str1" is an array).

In your printf(), "str1" could be either a pointer to a string or a character array (in which case the argument seen by printf() would be a pointer to the first element of the array).