Comments (2)
There's not a super-elegant way to do this right now, but basically yes you'd need to reimplement the standard operator logic inside your own handlers. Here's an example:
let expression = AnyExpression(parsed) { symbol in
switch symbol {
case .infix("<"):
return { args in
switch (args[0], args[1]) {
case let (lhs as Date, rhs as Date):
return lhs < rhs
case let (lhs as Double, rhs as Double):
return lhs < rhs
default:
throw Expression.Error.message("Type mismatch")
}
}
default:
return nil
}
}
from expression.
Thank you for the quick response! I’ll go with that for now, but would be nice to have a more elegant way down the track where the duplication wouldn’t be necessary.
from expression.
Related Issues (20)
- Usage for macOS Apps (Swift 5.1/Xcode 11.2.1) HOT 3
- Equality of strings, substrings and characters HOT 1
- Order of operations question HOT 3
- Square brackets HOT 1
- when using the sample code it won't execute HOT 4
- Expression.parse("'\\n'") fails HOT 2
- AnyExpression using array variables can't be evaluated with error 0 HOT 1
- Expose AST HOT 1
- Test suite crashes on Xcode 13.3 / Swift 5.6 HOT 7
- Ability to use ^ instead of pow(x, y) HOT 4
- Question: Can multiplication be done with parentheses? HOT 1
- Function expression failed with string paramater which contains \ HOT 1
- Crash in Swift 5.7 when creating stringBounds error HOT 4
- Getting sub expressions during evaluation HOT 1
- Lazy evaluation of ternary operator's true/false cases HOT 1
- Question: Is implicit multiplication possible with Expression? HOT 5
- Support ^ as pow operator HOT 2
- Fact `!` postfix operator does not work HOT 6
- Complex expressions are not working HOT 6
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from expression.