如何解决如何为具有可选参数的函数定义回调?
这是一个最小的例子:
import _Differentiation
@differentiable(reverse)
func g(x: Double?) -> Double {
if x == nil {
return 0.0 // I don't care what this value is
}
else {
return x! * x!
}
}
@derivative(of: g)
func gVJP(x: Double?) -> (value: Double,pullback: (Double) -> Double?) {
let value = g(x: x)
func pullback(_ dOutput: Double) -> Double? {
if x == nil {
return nil // I don't care what this value is
}
else {
return dOutput * 2.0 * x!
}
}
return (value: value,pullback: pullback)
}
我收到以下编译器错误:
Function result's 'pullback' type does not match 'g(x:)'
1. 'pullback' does not have expected type '(Double.TangentVector) -> Optional<Double>.TangentVector' (aka '(Double) -> Optional<Double>.TangentVector')
我尝试将回调的返回类型定义为 Optional<Double>.TangentVector
,但这给我的错误是我的回调的返回值为 incompatible with return type 'Optional<Double>.TangentVector'
。
解决方法
pullback
的返回值需要强制转换为类型 Optional<Double>.TangentVector
:
import _Differentiation
@differentiable(reverse)
func g(x: Double?) -> Double {
if x == nil {
return 0.0 // I don't care what this value is
}
else {
return x! * x!
}
}
@derivative(of: g)
func gVJP(x: Double?) -> (value: Double,pullback: (Double) -> Optional<Double>.TangentVector) {
let value = g(x: x)
func pullback(_ dOutput: Double) ->Optional<Double>.TangentVector {
if x == nil {
return Optional<Double>.TangentVector(nil) // I don't care what this value is
}
else {
return Optional<Double>.TangentVector(dOutput * 2.0 * x!)
}
}
return (value: value,pullback: pullback)
}
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