问题
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++ = valmaybe 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
回答1:
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.
回答2:
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++),
- first p++ evaluates, but the previous p value is returned
- then *p is evaluated with this previous p value and the assignment occurs
- then the statement ends and p++ post increment value is now active, ie pointer p is now bumped.
回答3:
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).
来源：https://stackoverflow.com/questions/15345396/precedence-of-dereference-and-postfix