我想使用 Swift 将字节从 uint8_t 数组转换为整数。
“C”示例:
char bytes[2] = {0x01, 0x02};
NSData *data = [NSData dataWithBytes:bytes length:2];
NSLog(@"data: %@", data); // data: <0102>
uint16_t value2 = *(uint16_t *)data.bytes;
NSLog(@"value2: %i", value2); // value2: 513
快速尝试:
let bytes:[UInt8] = [0x01, 0x02]
println("bytes: \(bytes)") // bytes: [1, 2]
let data = NSData(bytes: bytes, length: 2)
println("data: \(data)") // data: <0102>
let integer1 = *data.bytes // This fails
let integer2 = *data.bytes as UInt16 // This fails
let dataBytePointer = UnsafePointer<UInt16>(data.bytes)
let integer3 = dataBytePointer as UInt16 // This fails
let integer4 = *dataBytePointer as UInt16 // This fails
let integer5 = *dataBytePointer // This fails
在 Swift 中从 UInt8 数组创建 UInt16 值的正确语法或代码是什么?
我对 NSData 版本感兴趣,正在寻找不使用临时数组的解决方案。
最佳答案
如果你想通过 NSData
进行访问,那么它会像这样工作:
let bytes:[UInt8] = [0x01, 0x02]
println("bytes: \(bytes)") // bytes: [1, 2]
let data = NSData(bytes: bytes, length: 2)
print("data: \(data)") // data: <0102>
var u16 : UInt16 = 0 ; data.getBytes(&u16)
// Or:
let u16 = UnsafePointer<UInt16>(data.bytes).memory
println("u16: \(u16)") // u16: 513
或者:
let bytes:[UInt8] = [0x01, 0x02]
let u16 = UnsafePointer<UInt16>(bytes).memory
print("u16: \(u16)") // u16: 513
两种变体都假定字节是主机字节顺序。
Swift 3 (Xcode 8) 更新:
let bytes: [UInt8] = [0x01, 0x02]
let u16 = UnsafePointer(bytes).withMemoryRebound(to: UInt16.self, capacity: 1) {
$0.pointee
}
print("u16: \(u16)") // u16: 513
关于swift - 在 Swift 中将两个字节的 UInt8 数组转换为 UInt16,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25267089/