Does *p++ increment after dereferencing? [duplicate]

Answer 1

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.

Answer 2

Given q = *p++;, q gets the value that p pointed to before the increment. Another way to say it is that the value of the expression *p++ is the value that p pointed to before being incremented.

Answer 3

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.

Answer 4

The postfix ++ and -- operators essentially have higher precedence than the prefix unary operators. Therefore, *p++ is equivalent to *(p++); it increments p, and returns the value which p pointed to before p was incremented.

To increment the value pointed to by p, use (*p)++ (or perhaps ++*p, if the evaluation order of the side effect doesn't matter).

Answer 5

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.