Predicate

1.2.1

Treat predicates abstractly.
YOCKOW/SwiftPredicate

What's New

1.2.1

2022-10-05T09:31:22Z

What's Changed

  • Update dependency to suppor Swift 5.2 by @YOCKOW in #5

Full Changelog: 1.2.0...1.2.1

What is SwiftPredicate?

You can abstractly treat predicates with this library.

Requirements

  • Swift 5 (Also OK is compatibility mode for Swift 4 or 4.2)
  • macOS or Linux

Dependency

Usage

Simple Predicate

import Predicate

struct SimplePredicate<Variable>: PredicateProtocol {
  var _predicate: (Variable) -> Bool
  
  init(_ predicate:@escaping (Variable) -> Bool) {
    self._predicate = predicate
  }
  
  func evaluate(with argument: Variable) -> Bool {
    return self._predicate(argument)
  }
}

let lessThan10 = SimplePredicate<Int>({ $0 < 10 })
let greaterThan0 = SimplePredicate<Int>({ $0 > 0 })

// `PredicateProtocol` provides some operations like below:

print(lessThan10.and(greaterThan0).evaluate(with:5)) // Prints "true" 
print(lessThan10.and(greaterThan0).evaluate(with:-5)) // Prints "false"

print(lessThan10.or(greaterThan0).evaluate(with:15)) // Prints "true" 

print(lessThan10.xor(greaterThan0).evaluate(with:-5)) // Prints "true" 
print(lessThan10.xor(greaterThan0).evaluate(with:5)) // Prints "false"

A set defined by a predicate

There is also a set named "TotallyOrderedSet" that conforms to SetAlgebra and ConsolidatablePredicate (that inherits from PredicateProtocol). You can define elements contained by the set using ranges. (See SwiftRanges if you want to know what AnyRange is.)

import Predicate
import Ranges

let set1 = TotallyOrderedSet<Double>(elementsIn:[
  AnyRange<Double>(..<0.0),
  AnyRange<Double>(1.0...2.0),
  AnyRange<Double>(3.0<..)
])

let set2 = TotallyOrderedSet<Double>(elementsIn:[
  AnyRange<Double>((-2.0)<..(-1.0)),
  AnyRange<Double>(0.5..<1.5),
  AnyRange<Double>(2.0<..<3.5)
])

print(set1.inverted ==
      TotallyOrderedSet<Double>(elementsIn:[
        AnyRange<Double>(0.0..<1.0),
        AnyRange<Double>(2.0<..3.0),
      ])
     )
// Prints "true"

print(set1.intersection(set2) ==
      TotallyOrderedSet<Double>(elementsIn:[
        AnyRange<Double>((-2.0)<..(-1.0)),
        AnyRange<Double>(1.0..<1.5),
        AnyRange<Double>(3.0<..<3.5),
      ])
     )
// Prints "true"

print(set1.union(set2) ==
      TotallyOrderedSet<Double>(elementsIn:[
        AnyRange<Double>(..<0.0),
        AnyRange<Double>(0.5...),
      ])
     )
// Prints "true"

License

MIT License.
See "LICENSE.txt" for more information.

Description

  • Swift Tools 5.0.0
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Dependencies

Last updated: Fri Feb 03 2023 17:57:40 GMT-0500 (GMT-05:00)