Swift 4将字节转换为Int和Ntohl [英] Swift 4 convert bytes to Int and ntohl

问题描述

伙计们，所以我一直在尝试将这个Objective-C代码转换为快速代码，但始终会遇到麻烦。以下是目标C代码：

Hey guys so I've been trying to convert this objective-c code to swift but keep stumbling on problems with it. Heres the objective-c code:

int msgLength = *((int *) _inputBuffer.bytes);
msgLength = ntohl(msgLength);

，这里是我设法得到的东西：

and heres what I managed to get:

var msgLength = (inputBuffer.bytes).load(as: Int.self)
msgLength = Int(NSSwapBigLongToHost(UInt(msgLength)))

但这不起作用，它崩溃表示没有足够的位数。非常感谢您的帮助！

but this isn't working, it crashes saying there isn't enough bits. I really appreciate the help thanks!

推荐答案

C int 类型为一个32位整数（在当前所有Apple平台上）为
，而Swift Int 在64位平台上为64位整数。

The C int type is a 32-bit integer (on all current Apple platforms), whereas the Swift Int is 64-bit integer on 64-bit platforms.

因此，在Swift中，您必须对32位整数使用 UInt32 ：

Therefore in Swift you'll have to use UInt32 for 32-bit integers:

var msgLength = (inputBuffer.bytes).load(as: UInt32.self)
msgLength = UInt32(bigEndian: msgLength)

或者，如果您从 NSData 切换到 Data ：

Or, if you switch from NSData to Data:

let msgLength = UInt32(bigEndian:inputBuffer.withUnsafeBytes { $0.pointee })

（即使在C代码 uint32_t 中也更合适比 int 强调读取4个字节。）

(Even in the C code uint32_t would be better suited than int to emphasize that 4 bytes are read.)

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