What is the meaning of `printf(“%p”)` without arguments

Question

I of course know it used to output pointer with arguments.

I read book Writing Secure Code by Michael Howard and David LeBlanc.

One program in book

Answer 1

The behaviour of printf("Now the stack looks like:\n%p\n%p\n%p\n%p\n%p\n%p\n\n"); is undefined since your argument list does not match the input string.

The authors are conjecting that this particular misuse of printf will inject variables from the current stack, and output their addresses. Sometimes they might be correct. Other times the compiler might eat your cat.