Precedence of dereference and postfix

Question

When I read the TCPL by K&R, I just couldn't understand two expressions:

*p++ = val;  /*push val onto stack */

Here is my idea:

• dereference and postfix has the same precedence, and associativity is right to left,so

  *p++ = val maybe the same with *(p++) = val, because the pointer usually is the next position to the top , so in this code, p increase 1 first because of the parenthesis, so the p is the two units above the current top ,but not the one unit above the current top ,where the val should be!!! Thx

Answer 1

Precedence of operators is an order of their interpretation by compiler, not the order of their execution.

Operator precedence actually means "where to put parentheses". Hence you are correct that *p++ is the same as *(p++). But now we need to understand what is *(p++). It means taking *p and then increasing p++, because of post-fixed operation.

So, in short, you just mixed order of interpretation by compiler (which is determined by parentheses or precedence) and order of execution (which is determined by post- or pre-fixed definition).

Answer 2

The prefix increment/decrement and dereference operators are equal precedence, but the postfix operator is higher, so *p++ is the same as *(p++), which is like writing *p = val; p++;

If you wrote (*p)++ = val, it wouldn't compile, as you'd be trying to assign a value to a number.

Answer 3

Precedence and Associativity of Operators in K&R, table 2-1, pg 53, isn't as granular and complete as more recent table in Stroustrup, tC++PL,Sed, sec 6.2 Operator summary, p120-121.

C++ operator precedence Agnew's answer is excellent.

he points out association is indeed R->L for unary operators and that for *(p++),

1. first p++ evaluates, but the previous p value is returned
2. then *p is evaluated with this previous p value and the assignment occurs
3. then the statement ends and p++ post increment value is now active, ie pointer p is now bumped.