Why don't I need to dereference a character pointer in C before printing it?

Question

Why does this code work? I would expect that I would need to dereference ptr, printf(\"%s\\n\", *ptr); before I could print it out, but I get a

Answer 1

The format specifier %s tells printf to expect a pointer to a null-terminated char array. Which is what name and ptr are.

*name and *ptr are not. If you dereference them, you'll get back a single char, which is basically lying to printf - resulting in undefined behavior.