C Operator Precedence postfix increment and dereference

Question

Here is another naïve question from a C newbie: on this page, https://en.cppreference.com/w/c/language/operator_precedence, the precedence of the postfix increment is listed to be higher than that of pointer dereference. So I was expecting in the following code that the pointer is incremented first (pointing at 10) and then dereferenced.

#include <stdio.h>

int main()
{
  int a[] = {3, 10, 200};
  int *p = a;
  printf("%d", *p++);
  return 0;
}

However this code outputs still the first array item (3). What am I missing by the concept?

Answer 1

Precedence is placing of parenthesis.

The expression *p++ can be parenthesized as

(*p)++ // incorrect precedence
*(p++) // correct precedence

Note that the value of p++ is the value of p before any change, so the net effect of the correct precedence is the same as *p without ant reflection over the side-effect ++. The change to p itself does not alter the result of *(p++).

Answer 2

As you have correctly assumed, the expression *p++ is evaluated as *(p++); that is, the ++ operator has higher precedence than the * operator.

However, the value of the expression, p++, is just the value of p (i.e. its value before the increment). A side-effect of the operation is that the value of p is incremented after its value has been acquired.

From this Draft C11 Standard:

6.5.2.4 Postfix increment and decrement operators

…
2     The result of the postfix ++ operator is the value of the operand. As a side effect, the value of the operand object is incremented (that is, the value 1 of the appropriate type is added to it). … The value computation of the result is sequenced before the side effect of updating the stored value of the operand. With respect to an indeterminately-sequenced function call, the operation of postfix ++ is a single evaluation. …