線程1：致命錯誤：UnsafeMutablePointer.initialize重疊范圍

Question

我正在嘗試使用

UnsafeMutablePointer<UnsafeMutablePointer<Float>?>!

因為它是我需要使用的方法的參數所必需的。 但是我不知道這是什么或如何使用。

我通過執行以下操作創建了此值：

 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)

但是我得到了錯誤：

Thread 1: Fatal error: UnsafeMutablePointer.initialize overlapping range

我究竟做錯了什么？

Answer 1

您正在制造不良行為。

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

創建一個指向某些內存的指針，並將該內存初始化為存儲在bytes2的值。

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

---

然后，您決定創建一個引用該指針的內存的指針：

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

因此，如果您打印uint8Pointer ，它將具有與uint8Pointer2完全相同的值。如果您決定也更改其任何值，它也會更改uint8Pointer2的值。

因此，當您這樣做時：

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] uint8Pointer2覆蓋了uint8Pointer2的值。

---

到目前為止， uint8Pointer是只是一個淺拷貝uint8Pointer2 ..更改一個影響另一個..

現在，您決定執行以下操作：

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

在這里，您創建了一個指針（ uint8Pointer1 ）到uint8Pointer和你說uint8Pointer1初始化uint8Pointer ..但是你初始化的指針本身和8計數指針..

首先，不要打擾在指向具有自身值的指針的指針上調用initialize。它已經指向了正確的值。

很好的是：

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

將崩潰，但是：

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

不會..因為它不會對后者memcpy ，而前者卻不會。

希望我能正確解釋。

PS正確命名您的變量！

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C ++人士的翻譯：

//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.