Swift - 具有关联值和可交换性的模式匹配枚举
swift 5.0,Xcode 10.2.1
我在 Swift 中有一个 Expression
枚举。
enum Expression {
indirect case add(Expression, Expression)
indirect case subtract(Expression, Expression)
indirect case multiply(Expression, Expression)
indirect case divide(Expression, Expression)
indirect case power(Expression, Expression)
indirect case log(Expression, Expression)
indirect case root(Expression, Expression)
case x
case n(Int)
}
extension Expression: Equatable {
static func == (lhs: Expression, rhs: Expression) -> Bool {
switch (lhs, rhs) {
case let (.add(a, b), .add(c, d)) where a == c && b == d,
let (.subtract(a, b), .subtract(c, d)) where a == c && b == d,
let (.multiply(a, b), .multiply(c, d)) where a == c && b == d,
let (.divide(a, b), .divide(c, d)) where a == c && b == d,
let (.power(a, b), .power(c, d)) where a == c && b == d,
let (.log(a, b), .log(c, d)) where a == c && b == d,
let (.root(a, b), .root(c, d)) where a == c && b == d:
return true
case let (.n(a), .n(b)) where a == b:
return true
case (.x, .x):
return true
default:
return false
}
}
}
第一次尝试
我对我的 Expression
类型执行了很多模式匹配。加法和乘法的可交换性产生了很多长的模式匹配表达式。我想找到一种方法来简化和缩短它,所以我决定制作一个 ExpressionPattern
枚举并定义模式匹配运算符 (~=
) 的重载。
enum ExpressionPattern {
case commutativeMultiply(Expression, Expression)
case commutativeAdd(Expression, Expression)
}
func ~= (lhs: ExpressionPattern, rhs: Expression) -> Bool {
switch lhs {
case let .commutativeMultiply(a, b):
switch rhs {
case .multiply(a, b), .multiply(b, a):
return true
default:
return false
}
case let .commutativeAdd(a, b):
switch rhs {
case .add(a, b), .add(b, a):
return true
default:
return false
}
default:
return false
}
}
我希望能够替换模式匹配语句,例如:
case let .add(.n(3), a), let .add(a, .n(3)) where a > 10: //matches (a + 3), (3 + a)
//...
与:
case let .commutativeAdd(.n(3), a) where a > 10: //matches (a + 3), (3 + a)
//...
当我第一次尝试这样做时,我收到一条错误消息“模式变量绑定(bind)不能出现在表达式中”。
注意:我认为如果我使用没有值绑定(bind)的精确最终值,则此匹配有效,但我在整个项目的许多地方都使用了此功能。
尝试用法:
let expression: Expression = Expression.add(.divide(.n(1), .n(2)), .subtract(.n(3), .n(4)))
switch expression {
case let .commutativeAdd(.subtract(a, b), .divide(c, d)): //Error: Pattern variable binding cannot appear in an expression.
print("This matches: ((\(a) - \(b)) + (\(c) ÷ \(d))) and ((\(c) ÷ \(d) + (\(a) - \(b)))")
default:
break
}
第二次尝试
第二次尝试时,我更改了 Expression
的 ==
函数的定义,并尝试覆盖其默认实现的 ~=
.
extension Expression: Equatable {
static func == (lhs: Expression, rhs: Expression) -> Bool {
switch (lhs, rhs) {
case let (.add(a, b), .add(c, d)) where (a == c && b == d) || (a == d && b == c),
let (.subtract(a, b), .subtract(c, d)) where a == c && b == d,
let (.multiply(a, b), .multiply(c, d)) where (a == c && b == d) || (a == d && b == c),
let (.divide(a, b), .divide(c, d)) where a == c && b == d,
let (.power(a, b), .power(c, d)) where a == c && b == d,
let (.log(a, b), .log(c, d)) where a == c && b == d,
let (.root(a, b), .root(c, d)) where a == c && b == d:
return true
case let (.n(a), .n(b)) where a == b:
return true
case (.x, .x):
return true
default:
return false
}
}
static func ~= (lhs: Expression, rhs: Expression) -> Bool {
return lhs == rhs
}
}
尝试用法:
let expression: Expression = Expression.add(.divide(.n(1), .n(2)), .subtract(.n(3), .n(4)))
print(expression == .add(.subtract(.n(3), .n(4)), .divide(.n(1), .n(2)))) //Prints "true"
switch expression {
case .add(.subtract(.n(3), .n(4)), .divide(.n(1), .n(2))):
print("Matched")
default:
print("Not matched")
}
//Prints "Not matched"
注意:理想情况下应该打印“匹配”。
问题
我如何找到一种方法来使用成熟的模式匹配,同时考虑到某些情况的可交换性。
注意:这必须能够用于匹配具有可交换性的嵌套情况(即 .add(.multiply(..., ...), ...)
)。这一点尤其重要,因为在用加法和乘法匹配某些东西时需要更多的案例。
最佳答案
我改变了你的第二个用法如下
switch expression {
case let e where e == .add(.subtract(.n(3), .n(4)), .divide(.n(1), .n(2))):
print("Matched")
default:
print("Not matched")
}
我认为它会起作用。
关于Swift - 具有可交换性的模式匹配,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56431814/