这里有人可以看一下代码并告诉我它有什么问题吗?我基本上尝试构建一些通用函数,这些函数对某些原始类型(如 Int
、Float
、Double
等)进行操作。
不幸的是我无法让它正常工作。这是有效的代码(部分):
// 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
}
现在当我写的时候:
var doubleVec : [Double] = [1,2,3,4]
vec_dot(doubleVec, doubleVec)
它返回 30
的正确结果。好的,到目前为止一切顺利。当我尝试传递一组 Int
时,事情变得很奇怪:
extension Int : SummableMultipliable {}
var intVec : [Int] = [1,2,3,4]
vec_dot(intVec, intVec)
砰!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
我尝试执行不同的转换:
let x = Double(a[i] * b[1])
Error: Could not find an overload for 'init' that accepts the supplied arguments.
let y = a[i] * b[1]
let x = Double(y)
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
Error: Cannot invoke '*' with an argument list of type '(Double, Double').
我尝试了很多更多的东西。一旦我尝试将 Int
作为通用类型传递,就没有任何效果了。
也许我只是在这里遗漏了一些基本知识,或者我太笨了,无法理解泛型编程。在 C++ 中,我会在 2 秒内完成。
最佳答案
当用 Int
数组调用时,a[i] * b[i]
是一个 Int
并且不能被转换
到 Double
与 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 Double
。
要使初始化 var s : T = 0
起作用,您必须使 SummableMultipliable
派生自 IntegerLiteralConvertible
(Int
和 Double
已经符合):
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
}
例子:
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,如果向量积总是产生一个Double
,你可以
将 doubleValue()
方法添加到 SummableMultipliable
协议(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 所说的那样,循环应该写得更快
for i in 0 ..< a.count { ... }
如果你想炫耀你的同事并给他们留下深刻印象或迷惑,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 }
关于原始类型(Int、Float、Double)的泛型会产生奇怪的错误消息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25592838/