原始类型(Int，Float，Double)的泛型会产生奇怪的错误消息 [英] Generics of raw types (Int, Float, Double) create weird error messages

问题描述

这里有人可以看一下代码，然后告诉我这是怎么回事吗?我实质上是尝试构建一些对某些原始类型(例如Int，Float，Double等)运行的通用函数.

Can someone here maybe take a look at the code and tell me what's wrong with it? I essentially try to build a couple of generic functions that operate on certain raw types like Int, Float, Double etc.

很遗憾，我无法使其正常运行.这是部分有效的代码:

Unfortunately I can't get it working properly. This is the code that works (partially):

// 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30的正确结果.好的，到目前为止很好.当我尝试传递Int s数组时，事情变得很奇怪:

It returns the correct result of 30. Ok, so far so good. Things get weird when I try to pass an array of Ints:

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

Bam！引发异常:

 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

我尝试执行其他投射:

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

错误::找不到接受提供的参数的'init'的重载.

Error: Could not find an overload for 'init' that accepts the supplied arguments.

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'.

Error: Cannot invoke 'init' with an argument of type 'T'.

接下来，我尝试了:

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

错误:无法使用类型为((Double，Double))的参数列表调用'*'.

Error: Cannot invoke '*' with an argument list of type '(Double, Double').

我尝试了许多更多的东西.一旦我尝试将Int作为通用类型传递，就不再起作用了.

I tried out many more things. As soon as I try to pass Int as a generic type, nothing works anymore.

也许我只是在这里遗漏了一些基本知识，或者我太愚蠢而无法理解通用编程.在C ++中，我将在2秒内完成操作.

Maybe I'm just missing something fundamental here or I am just too dumb to understand generic programming. In C++, I'd be done in 2 seconds.

推荐答案

当使用Int数组调用时，a[i] * b[i]是Int，并且不能进行强制转换 到Double与as.

When called with an Int array, a[i] * b[i] is an Int and cannot be cast to Double with as.

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

To solve that problem, you can change your vec_dot function to return a T object instead of a Double. To make the initialization var s : T = 0 work, you have to make SummableMultipliable derive from IntegerLiteralConvertible (to which Int and Double already conform):

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
}

示例:

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，如果向量乘积应始终产生Double，则可以 在SummableMultipliable协议中添加doubleValue()方法:

---

Alternatively, if the vector product should always produce a Double, you can add a doubleValue() method to the SummableMultipliable protocol:

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正确说的那样，循环应更迅速地编写为

---

Remark: As @akashivskyy correctly said, the loop should be written more swiftly as

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

如果您想看中和打动或迷惑您的同事，则可以 用单个表达式替换整个循环:

If you want to get fancy and impress or puzzle your co-workers then you can replace the entire loop with a single expression:

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

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