线程 1:致命错误:不安全的可变指针初始化重叠范围

Question

>我正在尝试使用

UnsafeMutablePointer<UnsafeMutablePointer<Float>?>!

因为它是我需要使用的方法的参数所必需的。然而，我不知道这是什么或如何使用它。

我通过这样做创建了这个值：

var bytes2: [Float] = [39, 77, 111, 111, 102, 33, 39, 0]
let uint8Pointer2 = UnsafeMutablePointer<Float>.allocate(capacity: 8)
uint8Pointer2.initialize(from: &bytes2, count: 8)
var bytes: [Float] = [391, 771, 1111, 1111, 1012, 331, 319, 10]
var uint8Pointer = UnsafeMutablePointer<Float>?.init(uint8Pointer2)
uint8Pointer?.initialize(from: &bytes, count: 8)
let uint8Pointer1 = UnsafeMutablePointer<UnsafeMutablePointer<Float>?>!.init(&uint8Pointer)
uint8Pointer1?.initialize(from: &uint8Pointer, count: 8)

但是我收到错误：

Thread 1: Fatal error: UnsafeMutablePointer.initialize overlapping range

我做错了什么？

Answer 1

你正在制造不良行为。

var bytes2: [Float] = [39, 77, 111, 111, 102, 33, 39, 0]
let uint8Pointer2 = UnsafeMutablePointer<Float>.allocate(capacity: 8)
uint8Pointer2.initialize(from: &bytes2, count: 8)

创建指向某个内存的指针，并将该内存初始化为存储在bytes2中的值。

所以：uint8Pointer2 = [39, 77, 111, 111, 102, 33, 39, 0]

---

然后，您决定创建一个引用该指针内存的指针：

var uint8Pointer = UnsafeMutablePointer<Float>?.init(uint8Pointer2)

因此，如果您打印uint8Pointer，它将具有与uint8Pointer2完全相同的值。如果您决定也更改其任何值，它也会更改uint8Pointer2的值。

因此，当您这样做时：

var bytes: [Float] = [391, 771, 1111, 1111, 1012, 331, 319, 10]
uint8Pointer?.initialize(from: &bytes, count: 8)

它用[391, 771, 1111, 1111, 1012, 331, 319, 10]覆盖了uint8Pointer2的值..

---

到目前为止，uint8Pointer只是uint8Pointer2的浅拷贝..改变一个会影响另一个。

现在你决定做：

let uint8Pointer1 = UnsafeMutablePointer<UnsafeMutablePointer<Float>?>!.init(&uint8Pointer)
uint8Pointer1?.initialize(from: &uint8Pointer, count: 8)

在这里，你创建了一个指向uint8Pointer的指针(uint8Pointer1(，你说uint8Pointer1用uint8Pointer初始化..但是你正在初始化一个指针，指针指向自身，计数为8。

首先，不要费心调用初始化在指针到指针上，其值为自身。它已经指向正确的值。

好的是：

uint8Pointer1?.initialize(from: &uint8Pointer, count: 1)
//Same as:  memcpy(uint8Pointer1, &uint8Pointer, sizeof(uint8Pointer)`
//However, they both point to the same memory address..

会崩溃，但是：

uint8Pointer1?.initialize(from: &uint8Pointer)
//Same as: `uint8Pointer1 = uint8Pointer`.. Note: Just a re-assignment.

不会......因为它不会对后者做memcpy，而前者会。

希望我解释正确..

附言正确命名您的变量！

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C++人的翻译：

//Initial pointer to array..
float bytes2[] = {39, 77, 111, 111, 102, 33, 39, 0};
float* uint8Pointer2 = &bytes[2];
memcpy(uint8Pointer2, &bytes2[0], bytes2.size() * sizeof(float));
//Shallow/Shadowing Pointer...
float* uint8Pointer = uint8Pointer2;
float bytes[] = {391, 771, 1111, 1111, 1012, 331, 319, 10};
memcpy(uint8Pointer, &bytes[0], bytes.size() * sizeof(float));
//Pointer to pointer..
float** uint8Pointer1 = &uint8Pointer;
//Bad.. uint8Pointer1 and &uint8Pointer is the same damn thing (same memory address)..
//See the line above (float** uint8Pointer1 = &uint8Pointer)..
memcpy(uint8Pointer1, &uint8Pointer, 8 * sizeof(uint8Pointer));
//The memcpy is unnecessary because it already pointers to the same location.. plus it's also wrong lol.