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 %s formatter to printf expects a "string", which is really a pointer to a null-terminated array of characters.

That is to say, printf expects you to pass the pointer to it. When you dereference the pointer, it's taking the letter J, whose value in ascii is 74 (decimal), and tries treating that as a pointer to an array. That's an area of memory that's inaccessible, so you get the segmentation violation.