C＃中的变量拷贝到缓冲区中，而无需创建垃圾？ [英] C# Copy variables into buffer without creating garbage?

问题描述

是否有可能在C＃.NET（3.5及以上），以一个变量复制到byte []的缓冲区，而不在此过程中产生的任何垃圾？

Is it possible in C# .Net (3.5 and above) to copy a variable into a byte[] buffer without creating any garbage in the process?

例如：

int variableToCopy = 9861;

byte[] buffer = new byte[1024];
byte[] bytes = BitConverter.GetBytes(variableToCopy);
Buffer.BlockCopy(bytes, 0, buffer, 0, 4);

float anotherVariableToCopy = 6743897.6377f;
bytes = BitConverter.GetBytes(anotherVariableToCopy);
Buffer.BlockCopy(bytes, 0, buffer, 4, sizeof(float));

...

创建字节[]字节中介目标变成垃圾（presuming一个裁判不再追究的话）......

creates the byte[] bytes intermediary object which becomes garbage (presuming a ref is no longer held to it)...

我不知道如果使用位运算符的变量可以直接复制到缓冲区，而无需创建中介的byte []？

I wonder if using bitwise operators the variable can be copied directly into the buffer without creating the intermediary byte[]?

推荐答案

使用指针是最好的，最快的方法： 您可以使用任意数量的变量做到这一点，没有浪费的内存，固定语句有一个小的开销，但它太小了。

Use pointers is the best and the fastest way: You can do this with any number of variables, there is no wasted memory, the fixed statement has a little overhead but it's too small

        int v1 = 123;
        float v2 = 253F;
        byte[] buffer = new byte[1024];
        fixed (byte* pbuffer = buffer)
        {
            //v1 is stored on the first 4 bytes of the buffer:
            byte* scan = pbuffer;
            *(int*)(scan) = v1;
            scan += 4; //4 bytes per int

            //v2 is stored on the second 4 bytes of the buffer:
            *(float*)(scan) = v2;
            scan += 4; //4 bytes per float
        }

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