Does *p++ increment after dereferencing?

Try it. The program

#include <stdio.h>

int main(void) {
    int p[2];
    int *q = p;
    p[0] = 10;
    p[1] = 100;

    printf("%d\n", *q++);
    printf("%d\n", *q);

    return 0;
}

prints

10
100

showing that the ++ applies to p, not to *p, and that the increment happens after the dereference.

EDIT: (Thanks to @EricLippert for convincing me to pull out K & R)

Not only may there be a happens-after relationship, but according to K & R page 203, there must be:

A postfix expression followed by a ++ or -- operator is a postfix expression. The value of the expression of the expression is the value of the operand. After the value is noted, the operand is incremented (++) or decremented (--) by 1.

(emphasis mine)

Granted, I don't believe that K & R says anything about the semantics of C in the presence of multithreading (according to Wikipedia the pthreads specification was released in 1995), but for a single-threaded program K & R is pretty clear.

In the operators table, you can see that the suffix operator ++ have higher place than the * unary operator.

Hence, *p++ increase p (and not *p), and return the value of the address that p contained before the increment (since it's the suffix ++).

But the order is implementation-depend. It may begin by dereferencing p, and then increase it, and it may store the old value of p, increase it, and then dereference the old value.

There is no ordering between the increment and the dereference. However, the * operator applies to the result of p++, which is the original value of p prior to the increment.