Expert: Zlatko - 3/3/2010

Question
Why "&" is not required while reading a character array with "%s" specifier?

Answer
Hello Goraksh.

The & is not required because with %s you are reading the characters into an array. An array automatically evaluates to an address.

For example:

char str[123];
scanf("%s", str);

The str is automatically passed to scanf as the address of the start of the array.

Best regards
Zlatko