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

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 demonstrates how stack overflow works by strcpy()

Note printf() without arguments.

#include <stdio.h>
#include <string.h>

void foo(const char* input)
{
    char buf[10];

    //What? No extra arguments supplied to printf?
    //It's a cheap trick to view the stack 8-)
    //We'll see this trick again when we look at format strings.
    printf("My stack looks like:\n%p\n%p\n%p\n%p\n%p\n% p\n\n");

    //Pass the user input straight to secure code public enemy #1.
    strcpy(buf, input);
    printf("%s\n", buf);

    printf("Now the stack looks like:\n%p\n%p\n%p\n%p\n%p\n%p\n\n");
}

void bar(void)
{
    printf("Augh! I've been hacked!\n");
}

int main(int argc, char* argv[])
{
    //Blatant cheating to make life easier on myself
    printf("Address of foo = %p\n", foo);
    printf("Address of bar = %p\n", bar);
    if (argc != 2) 
    {
        printf("Please supply a string as an argument!\n");
        return -1;
        } 
    foo(argv[1]);
    return 0;
}

The result is

C:\Secureco2\Chapter05>StackOverrun.exe Hello
Address of foo = 00401000
Address of bar = 00401045
My stack looks like:
00000000
00000000
7FFDF000
0012FF80 
0040108A <-- return address
00410EDE

Hello
Now the stack looks like:
6C6C6548 <-- 'l','l','e','h'
0000006F <-- 0, 0, 0, 'o'
7FFDF000
0012FF80
0040108A
00410EDE

What is the meaning of printf("%p") inside code? Why it can print the content of stack?

via Chebli Mohamed