Parse an expression
An arithmetical expression consists of 2 numbers and an operator between them, for instance:
1 + 2
1.2 * 3.4
-3 / -6
-2 - 2
The operator is one of: "+"
, "-"
, "*"
or "/"
.
There may be extra spaces at the beginning, at the end or between the parts.
Create a function parse(expr)
that takes an expression and returns an array of 3 items:
- The first number.
- The operator.
- The second number.
For example:
let [a, op, b] = parse("1.2 * 3.4");
alert(a); // 1.2
alert(op); // *
alert(b); // 3.4
A regexp for a number is: -?\d+(\.\d+)?
. We created it in the previous task.
An operator is [-+*/]
. The hyphen -
goes first in the square brackets, because in the middle it would mean a character range, while we just want a character -
.
The slash /
should be escaped inside a JavaScript regexp /.../
, we’ll do that later.
We need a number, an operator, and then another number. And optional spaces between them.
The full regular expression: -?\d+(\.\d+)?\s*[-+*/]\s*-?\d+(\.\d+)?
.
It has 3 parts, with \s*
between them:
-?\d+(\.\d+)?
– the first number,[-+*/]
– the operator,-?\d+(\.\d+)?
– the second number.
To make each of these parts a separate element of the result array, let’s enclose them in parentheses: (-?\d+(\.\d+)?)\s*([-+*/])\s*(-?\d+(\.\d+)?)
.
In action:
The result includes:
result[0] == "1.2 + 12"
(full match)result[1] == "1.2"
(first group(-?\d+(\.\d+)?)
– the first number, including the decimal part)result[2] == ".2"
(second group(\.\d+)?
– the first decimal part)result[3] == "+"
(third group([-+*\/])
– the operator)result[4] == "12"
(forth group(-?\d+(\.\d+)?)
– the second number)result[5] == undefined
(fifth group(\.\d+)?
– the last decimal part is absent, so it’s undefined)
We only want the numbers and the operator, without the full match or the decimal parts, so let’s “clean” the result a bit.
The full match (the arrays first item) can be removed by shifting the array result.shift()
.
Groups that contain decimal parts (number 2 and 4) (.\d+)
can be excluded by adding ?:
to the beginning: (?:\.\d+)?
.
The final solution:
As an alternative to using the non-capturing ?:
, we could name the groups, like this: