我想访问现有的UInt64数组,就像它是一个Int8数组一样。关键要求是效率 - 我不想复制或重新分配数据,只需要直接访问。我不想要副作用(例如,我希望能够在执行此代码块之后继续使用uint64Array,正在阅读有关具有未定义副作用的重新绑定。)
我试过用Swift 4.2做这个:
var uint64Array = [UInt64](repeating: 0, count: 100)
uint64Array.withUnsafeMutableBufferPointer() {
uint64Pointer in
uint64Pointer.withMemoryRebound(to: Int8.self) { // <- Error occurs here.
int8Pointer in
int8Pointer[0] = 1
int8Pointer[1] = 2
int8Pointer[2] = 3
int8Pointer[3] = 4
}
}
但是,我在运行时在以下行中收到致命错误:
uint64Pointer.withMemoryRebound(to: Int8.self) {
这是正确的方法吗?如果是这样,为什么我会收到致命错误?
我认为问题是你不能直接按照文档中的这个注释绑定到不同的类型:
仅使用此方法将缓冲区的内存重新绑定到具有与当前绑定的Element类型相同大小和步幅的类型。要将内存区域绑定到不同大小的类型,请将缓冲区转换为原始缓冲区并使用bindMemory(to :)方法。
如果字节是您所追求的那么最快的路线是:
var uint64Array = [UInt64](repeating: 0, count: 100)
uint64Array.withUnsafeMutableBytes { x in
x[0] = 1
x[1] = 2
x[3] = 3
x[4] = 4
}
如果您有其他类型,您可以这样做:
var uint64Array = [UInt64](repeating: 0, count: 100)
uint64Array.withUnsafeMutableBufferPointer() {
uint64Pointer in
let x = UnsafeMutableRawBufferPointer(uint64Pointer).bindMemory(to: Int32.self)
x[0] = 1
x[1] = 2
x[3] = 3
x[4] = 4
}
感谢@brindy解决这个问题。这是一个尽可能干净的扩展实现。
扩展名:
extension Array {
mutating func bindMutableMemoryTo<T,R>(_ type: T.Type, _ closure: (UnsafeMutableBufferPointer<T>) throws -> R) rethrows -> R {
return try self.withUnsafeMutableBytes() {
return try closure($0.bindMemory(to: type))
}
}
}
用法:
var uint64Array = [UInt64](repeating: 0, count: 100)
uint64Array.bindMutableMemoryTo(Int8.self) {
int8Pointer in
int8Pointer[0] = 1 // LSB of uint64Array[0]
int8Pointer[1] = 2
int8Pointer[2] = 3
int8Pointer[3] = 4 // MSB of uint64Array[0]
}