Operator precedence is why the expression 5 + 3 * 2 is
calculated as 5 + (3 * 2), giving 11, and not as (5 + 3)
* 2, giving 16.
We say that the multiplication operator (*) has higher
"precedence" or "priority" than the addition operator (+), so
the multiplication must be performed first.
Operator associativity is why the expression 8  3  2 is
calculated as (8  3)  2, giving 3, and and not as 8 
(3  2), giving 7.
We say that the subtraction operator () is "left
associative", so the left multiplication must be performed first. When
we can't decide by operator precedence alone in which order to
calculate an expression, we must use associativity.
Since the operators + and  have the same
precedence, and are leftassociative, the following expressions are
all equivalent:
x + 3  y + 5
(x + 3)  y + 5
((x + 3)  y) + 5
((x + 3)  y + 5)
Since the operators =, += and = have
the same precedence, and are rightassociative, the following
expressions are all equivalent:
x = y += z = 4
x = y += (z = 4)
x = (y += (z = 4))
(x = y += (z = 4))
You can use parentheses to tell the compiler in which order to
evaluate things. If you don't use parentheses, Pike will use the
precedences and the associativities of the operators to decide in
which order to perform them. Spaces have no effect.
This table shows the priority and associativity of each operator in
Pike, with the highest priority at the top:
Operators  Associativity 
::a a::b

left to right

a() a[b] a>b a[b..c] ({}) ([]) (<>)

left to right

a++ a

left to right

!a ~a (type)a ++a a

right to left!

a*b a/b a%b

left to right

a+b ab

left to right

a>>b a<<b

left to right

a>b a>=b a<b a<=b

left to right

a==b a!=b

left to right

a&b

left to right

a^b

left to right

ab

left to right

&&

left to right



left to right

a?b:c

right to left!

=
+=
=
*=
/=
%=
<<=
>>=
&=
=
^=

right to left!

@a

right to left!

,

left to right

Examples:
This expression  is evaluated in this order 
1+2*2

1+(2*2)

1+2*2*4

1+((2*2)*4)

(1+2)*2*4

((1+2)*2)*4

1+4,c=23+5

(1+4),(c=(2(3+5)))

1 + 5&4 == 3

(1 + 5) & (4 == 3)

c=1,99

(c=1),99

!a++ + ~f()

(!(a++)) + (~(f()))

s == "klas"  i < 9

(s == "klas")  (i < 9)

r = s == "sten"

r = (s == "sten")
