Assume the following rules of associativity and precedence for expressions:

precedence
highest
- (unary), (prefix) ++, (prefix) --
(postfix) ++, (postfix) --
+, *, & , mod
-, /, not
< , =, > , ! =
=, +=, *=, /=

and
or, xor
lowest
associativity
left to right

show the order of evaluation of the following expressions by parenthesizing all subexpressions and placing a superscript on the right parenthesis to indicate order. for example, for the expression a * b + c / d the order of evaluation would be represented as

((a * b)ˆ1 + c)ˆ2 / d)ˆ3

also rewrite the expression where there are no precedence rules and the statement is given right to left associativity, for example for the same problem above

d / c + b * a

a. a * b - 1 + c
b. a * (b - 1) / (++c++) mod d
c. (a - b) / c & (d * e / a - 3)
d. -a or c = d and e
e. a > b xor c or d < = 17
f. -a + b

The answers are explained below.

Explanation:

a.) * and + has same priority so then: ((a * b) - (1+c))

b.) For (++c++) = ((++c)++) , prefix has high precedence.

In remaining operators * and mod has same precedence So evaluate from left to right. ((a * (b - 1)) / (((++c)++) mod d))

There for order of evaluation: ((a * (b - 1)) / (((++c)++) mod d))

c.) for (d * e / a - 3), * has high priority so ((d *e) / a - 3). Now / and - have same priority so then see association from left to right. so

And & has more priority than / so then order of evaluation: ((a - b) / (c & (((d * e) / a) - 3))

d.) In question d, unary minus(-) has high priority then, - has high priority, then and has high priority and then or has low priority. So then order of evaluation: ((-a) or ((c = d) and e))

e.) Here > and <= has high and same precedence then ( a > b) xor c or (d <= 17)

now xor and or has same priority so associates from left to right.

order of evaluation is : ( ((a > b) xor c) or (d <= 17))

f.) uniary minus(-) has high precedence than + so order of evaluation is : ((-a) + b)

Rewriting the expression where there are no precedence rues and the statement is given right to left associativity:

a. c + 1 - b * a

b. d mod (++c++) / (1 - b) * b

c. (3 - a / e * d) & c / (b - a)

d. e and d = c or -a

e. 17 <= d or c xor b > a

f. b + -a

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