Getting DDMathParser tokens and group tokens

Question

I already found a solution to my problem on stackoverflow but it is in Objective-C. See this link DDMathParser - Getting tokens

I translated this into Swift as shown below. So what is the latest way to get tokens from a string and to get grouped tokens?

For example: 1 + ($a - (3 / 4))

Here is my try:

do{
    let token = try Tokenizer(string: "1 + ($a - (3 /4))").tokenize()
    for element in token {
        print(element)
    }
} catch {
    print(error)
}

But I get the following messages:

MathParser.DecimalNumberToken
MathParser.OperatorToken
MathParser.OperatorToken
MathParser.VariableToken
MathParser.OperatorToken
MathParser.OperatorToken
MathParser.DecimalNumberToken
MathParser.OperatorToken
MathParser.DecimalNumberToken
MathParser.OperatorToken
MathParser.OperatorToken

How do I get the specific tokens from my string?

Could u look at my code and tell me what I am missing, please ? — TBH, Nov 17 '17 at 13:32
*How do I get the specific tokens from my string ?* You are iterating over the tokens, so you have them already. — vikingosegundo, Nov 17 '17 at 14:40
@DaveDeLong What i mean by "grouped" tokens are the group terms. Here in my example we would have 3 group terms. My question is how do I get the group terms ? — TBH, Nov 18 '17 at 21:25
@TBH in that case, I think you may just want the final `Expression` tree. — Dave DeLong, Nov 18 '17 at 21:25

Dave DeLong · Accepted Answer · 2017-11-18T21:56:49.497

0

You're getting all of the tokens.

OperatorToken, DecimalNumberToken, etc all inherit from a RawToken superclass. RawToken defines a .string and a .range property.

The .string is the actual String, as extracted or inferred from the source. The .range is where in the original string the token is located.

It's important to note, however, that there may be tokens produced by the tokenizer that are not present in the original string. For example, tokens get injected by the Tokenizer when resolving implicit multiplication (3x turns in to 3 * x).

Added later:

If you want the final tree of how it all gets parsed, then you want the Expression:

let expression = try Expression(string: "1 + ($a - (3 / 4))")

At this point, you switch on expression.kind. It'll either be a number, a variable, or a function. function expressions have child expressions, representing the arguments to the function.

edited Nov 18 '17 at 21:56

answered Nov 18 '17 at 01:58

Dave DeLong

242,470
58
448
498

If I have the following expression "let expression = try Expression(string: "1 - ($a - (3/4)) + ($a + 2)"), I get the message "function("add", [1.0 - $a - 3.0 / 4.0 + $a, 2.0])". My question: How do I know that $a-3/4 is in brackets ? – TBH Nov 29 '17 at 12:39
@TBH that looks like the `description`. As I mentioned, you'll want to `switch` on the `expression.kind`. That'll tell you it's a `.function` expression, and you'll get the 2 arguments as part of the enum payload. The two arguments will both be expressions: the first will be `.number(1.0)`, and the second will be what you're looking for – Dave DeLong Nov 30 '17 at 00:24

Getting DDMathParser tokens and group tokens

1 Answers1