原始类型(Int,Float,Double)的泛型创建了奇怪的错误消息

Question

这里有人可以看看代码并告诉我它有什么问题吗？我基本上是试图构建一对夫妇的是对某些原始类型,如操作的通用功能Int,Float,Double等.

不幸的是我无法让它正常工作.这是(部分)工作的代码:

// http://stackoverflow.com/a/24047239/2282430
protocol SummableMultipliable: Equatable {
    func +(lhs: Self, rhs: Self) -> Self
    func *(lhs: Self, rhs: Self) -> Self
}

extension Double: SummableMultipliable {}

func vec_dot<T where T: SummableMultipliable>(a : [T], b: [T]) -> Double {
    assert(a.count == b.count, "vectors must be of same length")
    var s : Double = 0.0
    for var i = 0; i < a.count; ++i {
        let x = (a[i] * b[i]) as Double
        s = s + x
    }
    return s
}

现在我写的时候:

 var doubleVec : [Double] = [1,2,3,4]

 vec_dot(doubleVec, doubleVec)

它返回正确的结果30.好的,到目前为止一切顺利.当我尝试传递一个Ints 数组时,事情变得奇怪:

 extension Int : SummableMultipliable {}
 var intVec : [Int] = [1,2,3,4]
 vec_dot(intVec, intVec)

巴姆!抛出异常:

 let x = (a[1] * b[1]) as Double

* thread #1: tid = 0x139dd0, 0x00000001018527ad libswiftCore.dylib`swift_dynamicCast + 1229, queue = 'com.apple.main-thread', stop reason = EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0)
  * frame #0: 0x00000001018527ad libswiftCore.dylib`swift_dynamicCast + 1229
    frame #1: 0x000000010d6c3a09 $__lldb_expr248`__lldb_expr_248.vec_dot <A : __lldb_expr_248.SummableMultipliable>(a=Swift.Array<T> at 0x00007fff5e5a9648, b=Swift.Array<T> at 0x00007fff5e5a9640) -> Swift.Double + 921 at playground248.swift:54
    frame #2: 0x000000010d6c15b0 $__lldb_expr248`top_level_code + 1456 at playground248.swift:64
    frame #3: 0x000000010d6c4561 $__lldb_expr248`main + 49 at <EXPR>:0
    frame #4: 0x000000010165b390 FirstTestPlayground`get_field_types__XCPAppDelegate + 160
    frame #5: 0x000000010165bea1 FirstTestPlayground`reabstraction thunk helper from @callee_owned () -> (@unowned ()) to @callee_owned (@in ()) -> (@out ()) + 17
    frame #6: 0x000000010165ab61 FirstTestPlayground`partial apply forwarder for reabstraction thunk helper from @callee_owned () -> (@unowned ()) to @callee_owned (@in ()) -> (@out ()) + 81
    frame #7: 0x000000010165bed0 FirstTestPlayground`reabstraction thunk helper from @callee_owned (@in ()) -> (@out ()) to @callee_owned () -> (@unowned ()) + 32
    frame #8: 0x000000010165bf07 FirstTestPlayground`reabstraction thunk helper from @callee_owned () -> (@unowned ()) to @callee_unowned @objc_block () -> (@unowned ()) + 39
    frame #9: 0x0000000101fedaac CoreFoundation`__CFRUNLOOP_IS_CALLING_OUT_TO_A_BLOCK__ + 12
    frame #10: 0x0000000101fe37f5 CoreFoundation`__CFRunLoopDoBlocks + 341
    frame #11: 0x0000000101fe2fb3 CoreFoundation`__CFRunLoopRun + 851
    frame #12: 0x0000000101fe29f6 CoreFoundation`CFRunLoopRunSpecific + 470
    frame #13: 0x000000010208f2b1 CoreFoundation`CFRunLoopRun + 97
    frame #14: 0x0000000101658be8 FirstTestPlayground`top_level_code + 3784
    frame #15: 0x000000010165b3ba FirstTestPlayground`main + 42
    frame #16: 0x0000000103cd9145 libdyld.dylib`start + 1

我尝试执行不同的演员:

let x = Double(a[i] * b[1])

错误:找不到接受提供的参数的'init'的重载.

let y = a[i] * b[1]
let x = Double(y)

错误:无法使用类型为"T"的参数调用"init".

接下来,我试过:

let y = Double(a[i]) * Double(b[1])
let x = y

错误:无法使用类型'(Double,Double')的参数列表调用'*'.

我尝试了很多更多的东西.一旦我尝试Int作为通用类型传递,任何东西都不再有效.

也许我只是缺少一些基本的东西,或者我太愚蠢了解通用编程.在C++中,我将在2秒内完成.

Answer 1

当用Int数组调用时,a[i] * b[i]是一个Int并且不能被强制转换Double为as.

要解决该问题,您可以更改vec_dot函数以返回T 对象而不是a Double.要使初始化var s : T = 0工作,你必须SummableMultipliable 从IntegerLiteralConvertible(到哪个Int并且Double已经符合)派生:

protocol SummableMultipliable: Equatable, IntegerLiteralConvertible {
    func +(lhs: Self, rhs: Self) -> Self
    func *(lhs: Self, rhs: Self) -> Self
}

func vec_dot<T where T: SummableMultipliable>(a : [T], b: [T]) -> T {
    assert(a.count == b.count, "vectors must be of same length")
    var s : T = 0
    for var i = 0; i < a.count; ++i {
        let x = (a[i] * b[i])
        s = s + x
    }
    return s
}

例:

var doubleVec : [Double] = [1,2,3,4]
let x = vec_dot(doubleVec, doubleVec)
println(x) // 30.0 (Double)
var intVec : [Int] = [1,2,3,4]
let y = vec_dot(intVec, intVec)
println(y) // 30 (Int)

或者,如果矢量产品应始终生成a Double,则可以doubleValue()向SummableMultipliable协议添加方法:

protocol SummableMultipliable: Equatable {
    func +(lhs: Self, rhs: Self) -> Self
    func *(lhs: Self, rhs: Self) -> Self
    func doubleValue() -> Double
}

extension Double: SummableMultipliable {
    func doubleValue() -> Double { return self }
}

extension Int : SummableMultipliable {
    func doubleValue() -> Double { return Double(self) }
}

func vec_dot<T where T: SummableMultipliable>(a : [T], b: [T]) -> Double {
    assert(a.count == b.count, "vectors must be of same length")
    var s : Double = 0
    for var i = 0; i < a.count; ++i {
        let x = (a[i] * b[i]).doubleValue()
        s = s + x
    }
    return s
}

备注:正如@akashivskyy正确地说的那样,循环应该写得更快

for i in 0 ..< a.count { ... }

如果你想获得幻想,给同事留下深刻印象或困惑,那么你可以用一个表达式替换整个循环:

let s : T = reduce(Zip2(a, b), 0) { $0 + $1.0 * $1.1 }