英文:
ByteBuffer.wrap().getInt() equivalent in c#
问题
**Java**
byte[] input = new byte[] { 83, 77, 45, 71, 57, 51, 53, 70 };
int buff = ByteBuffer.wrap(input).getInt();
输出：`1397566791`
**C＃**
byte[] array = { 83, 77, 45, 71, 57, 51, 53, 70 };
MemoryStream stream = new MemoryStream();
using (BinaryWriter writer = new BinaryWriter(stream))
{
writer.Write(array);
}
byte[] bytes = stream.ToArray();
int buff = BitConverter.ToInt32(bytes, 0);
输出：`1194151251`
我不知道如何获得*相同的*输出
谢谢
英文:
Java
byte[] input = new byte[] { 83, 77, 45, 71, 57, 51, 53, 70 };
int buff = ByteBuffer.wrap(input).getInt();
Output: 1397566791
C#
byte [] array = { 83, 77, 45, 71, 57, 51, 53, 70 };
MemoryStream stream = new MemoryStream();
using (BinaryWriter writer = new BinaryWriter(stream))
{
writer.Write(array);
}
byte[] bytes = stream.ToArray();
int buff = BitConverter.ToInt32(bytes, 0);
Output: 1194151251
I have no idea how to get the same output
Thanks
答案1
得分: 2
好的，以下是翻译好的部分：
好的，`Int32` 仅包含 `4` 个字节，让我们通过 `Take(4)` 来获取它们的帮助。接下来，我们必须考虑 *结束*（Big 或 Little）并在必要时`Reverse`这些 `4` 个字节：
using System.Linq;
...
byte[] array = { 83, 77, 45, 71, 57, 51, 53, 70 };
// 1397566791
int buff = BitConverter.ToInt32(BitConverter.IsLittleEndian
? array.Take(4).Reverse().ToArray()
: array.Take(4).ToArray());
英文:
Well, Int32 consists of 4 bytes only, let's Take them with the help of Take(4). Next, we have to take ending (Big or Little) into account and Reverse these 4 bytes if necessary:
using System.Linq;
...
byte[] array = { 83, 77, 45, 71, 57, 51, 53, 70 };
// 1397566791
int buff = BitConverter.ToInt32(BitConverter.IsLittleEndian
? array.Take(4).Reverse().ToArray()
: array.Take(4).ToArray());
答案2
得分: 1
在Java情况下，它会按顺序取前4个字节并将它们转换为int。
System.out.println((((((83<<8)+77)<<8)+45)<<8)+71);
1397566791
在C#中，它会按相反的顺序取前四个字节。
System.out.println((((((71<<8)+45)<<8)+77)<<8)+83);
1194151251
因此，您需要阅读描述两个类操作的API文档，并相应地使用它们。应该有一种方法可以反转字节顺序。
从C#转到Java，可能会像这样。
buff = ByteBuffer.wrap(input).order(ByteOrder.LITTLE_ENDIAN).getInt();
System.out.println(buff);
输出
1194151251
英文:
In the Java case, it's taking the first 4 bytes in order and converting them to an int.
System.out.println((((((83<<8)+77)<<8)+45)<<8)+71);
1397566791
In C# it is taking the first four in reverse order.
System.out.println((((((71<<8)+45)<<8)+77)<<8)+83);
1194151251
So you need to read the API documentation that describes the operation for both classes and use them accordingly. There should be a way to reverse the byte order.
To go from C# to Java it would be something like this.
buff = ByteBuffer.wrap(input).order(ByteOrder.LITTLE_ENDIAN).getInt();
System.out.println(buff);
Prints
1194151251
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