# Operator precedence and associativity¶

When we write something like \(1 + 2 \times 3\), how do we know what it means? Does it mean \((1 + 2) \times 3\), or \(1 + (2 \times 3)\)? Of course, you are familiar with the usual “order of operations”, where multiplication comes before addition, so in fact \(1 + 2 \times 3\) should be interpreted as \(1 + (2 \times 3) = 1 + 6 = 7\).

Another way to say this is that multiplication has *higher precedence*
than addition. If we think of operators as “magnets” that attract
operands, higher precedence operators are like “stronger magnets”.

Another issue arises when operators are repeated, or when operators
with the same precedence are used together. For example, does
\(4 - 3 - 2 - 1\) mean \(((4 - 3) - 2) - 1\) or \(4 -
(3 - (2 - 1))\)? In fact, it means the former, because addition and
subtraction are done “left to right”; we say they are *left
associative*. On the other hand, exponentiation is *right associative*,
meaning that `1 ^ 2 ^ 3 ^ 4 = 1 ^ (2 ^ (3 ^ 4))`

.

Every operator in Disco has a precedence level and associtivity, and
Disco uses these to determine where to put parentheses in
expressions like `1 + 2 * 3`

or ```
5 > 2 ^ 2 + 1 and 7 > 2 ==>
true
```

. You might have memorized something like PEMDAS, but Disco has
so many operators that memorizing their precedence levels is out of
the question! Instead, we can use the `:doc`

command to show us the
precedence level and associativity of different operators.

```
Disco> :doc ^
~^~ : ℕ × ℕ → ℕ
precedence level 13, right associative
Disco> :doc +
~+~ : ℕ × ℕ → ℕ
precedence level 7, left associative
Disco> :doc >
~>~ : ℕ × ℕ → Bool
precedence level 5, right associative
Disco> :doc and
~and~ : Bool × Bool → Bool
precedence level 4, right associative
Disco> :doc ==>
~==>~ : Bool × Bool → Bool
precedence level 2, right associative
```