C中的运算符优先级

Question

have a trouble understanding this expression:

(x + y - 1) / y * y

The operators precedence is as follows (as per my understanding and K&R2, 2.12 table 2.1):

1) evaluate what is in the parens: firstly (x+y), then (x+y) -1

2) '*' operator has higher priority then '/' so it's supposed to go first, bit it appears that (y * y) is evaluated and then the result of (x+y-1) is divided by the product (y*y). I don't quite get it.

3) I ever heard that normally rounding up iw written in this form:

(x + y - 1) / y * y

Is that correct? Thank you very much in advance!

Answer 1

Unary operator * has higher precedence than / , but that is for pointer dereferencing.

Multiplication * and division / have the same left-to-right precedence.

Answer 2

no, * and / have same precedence. (it is called " precedence ", not "priority")

(x + y - 1) / y * y

would be

( ( (x+y) - 1 ) / y ) * y

The operation tree would be:

           *
          / \
         /   y
        ÷
       / \
      /   y
     -
    / \
   /   1
  +
 / \
x   y

Answer 3

As mentioned earlier the "*" and "/" have the same precedence so they are evaluated left to right. Totally disambiguating the expression gives:

( ( ( (x + y) - 1) / y) * y)

Answer 4

1) right

2) No, "/" and "*" is the same priority. So it will be performed from left to right.

3) I don't understand what your "round up" mean. But for example:

With y = 2, x = 2

(x + y -1) / y * y = (2 + 2 - 1) / 2 * 2 = (3 / 2) * 2 = 1 * 2 = 2

3 / 2 = 1 because this is the division of integer.

Answer 5

'*' operator has higher priority then '/'

Incorrect. Both * and / have equal precedence and are left-associative.

After evaluating the parenthesis expression we have:

(x + y - 1) / y * y = z / y * y      // where z = x + y -1

                    = (z / y) * y    // because of associativity.

Answer 6

I may be wrong, but as far as I recall / and * have same precedence. So the result should be (x+y+1).

Maybe you should say (x+y+1)/(y*y) if that's what you want the result to be.

It never costs much to disambiguate in any case.

As to using ((x+y-1)/y)*y to round up or down as the case may be, this is a dubious and not very portable practice. Several reasons: 1/ you must know or remember that y is some sort of int, 2/ x must also be some sort of int, 3/ results may vary depending on the compiler.