ios - 将 UnsafeMutablePointer<UInt8> 转换为字符串时堆缓冲区溢出

Question

我在执行以下操作时从 Address Sanitizer 收到错误消息:

let pointer = UnsafeMutableRawPointer.allocate(byteCount: 4, alignment: 1)
pointer.storeBytes(of: 77, as: UInt8.self)
pointer.advanced(by: 1).storeBytes(of: 105, as: UInt8.self)
(pointer+2).storeBytes(of: 107, as: UInt8.self)
(pointer+3).storeBytes(of: 101, as: UInt8.self)

let typedPointer = pointer.bindMemory(to: UInt8.self, capacity: 4)


let readableData = String(cString: typedPointer)

我得到堆缓冲区溢出，但我不明白为什么。在我的实际代码中，我有一个复杂得多的指针，但即使在这个简单的示例中，我也一直遇到这个问题。我认为它与 String(cString: typedPointer) 有关，但我没有看到我如何分配错误的内存大小，这会导致任何堆 header 或数据被踩踏。

更新 - 请参阅下面的答案

看起来我需要一个空终止符作为指针中的最后一个字节，否则 String 将不知道指针在哪里结束。

Answer 1

其他选项...

您可以从您的 UnsafeMutableRawPointer 创建Data:

let pointer = UnsafeMutableRawPointer.allocate(byteCount: 4, alignment: 1)
pointer.storeBytes(of: 77, as: UInt8.self)
pointer.advanced(by: 1).storeBytes(of: 105, as: UInt8.self)
(pointer+2).storeBytes(of: 107, as: UInt8.self)
(pointer+3).storeBytes(of: 101, as: UInt8.self)

let data = Data(bytes: pointer, count: 4)
let readableData = String(data: data, encoding: .utf8)

否则，String.init(bytes:encoding:) 是另一个不声明空终止序列的初始化器:

let pointer = UnsafeMutableRawPointer.allocate(byteCount: 4, alignment: 1)
pointer.storeBytes(of: 77, as: UInt8.self)
pointer.advanced(by: 1).storeBytes(of: 105, as: UInt8.self)
(pointer+2).storeBytes(of: 107, as: UInt8.self)
(pointer+3).storeBytes(of: 101, as: UInt8.self)

let urbp = UnsafeRawBufferPointer(start: pointer, count: 4)
let readableData = String(bytes: urbp, encoding: .utf8)

请尝试。