arrays - 在没有复制或分配的 Swift 数组类型之间进行转换

Question

我想访问一个现有的 UInt64 数组，就好像它是一个 Int8 数组一样。关键要求是效率——我不想复制或重新分配数据，只想直接访问。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(bind)。)

我试着用 Swift 4.2 来做这个:

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遇到 fatal error :

    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 fatal error ？

Answer 1

我认为问题在于您不能按照文档中的注释直接绑定(bind)到不同的类型:

Only use this method to rebind the buffer’s memory to a type with the same size and stride as the currently bound Element type. To bind a region of memory to a type that is a different size, convert the buffer to a raw buffer and use the bindMemory(to:) method.

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

}

如果您想使用其他类型，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

}