You can abstractly treat predicates with this library.
- Swift 5 (Also OK is compatibility mode for Swift 4 or 4.2)
- macOS or Linux
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"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"MIT License.
See "LICENSE.txt" for more information.