如何有效地保留填充字节数组

Question

假设我有一个数组

LogoDataBy
{byte[0x00000008]}
    [0x00000000]: 0x41
    [0x00000001]: 0x42
    [0x00000002]: 0x43
    [0x00000003]: 0x44
    [0x00000004]: 0x31
    [0x00000005]: 0x32
    [0x00000006]: 0x33
    [0x00000007]: 0x34

我想创建一个任意长度的数组,并将其填充 0x00

newArray
{byte[0x00000010]}
    [0x00000000]: 0x00
    [0x00000001]: 0x00
    [0x00000002]: 0x00
    [0x00000003]: 0x00
    [0x00000004]: 0x00
    [0x00000005]: 0x00
    [0x00000006]: 0x00
    [0x00000007]: 0x00
    [0x00000008]: 0x41
    [0x00000009]: 0x42
    [0x0000000a]: 0x43
    [0x0000000b]: 0x44
    [0x0000000c]: 0x31
    [0x0000000d]: 0x32
    [0x0000000e]: 0x33
    [0x0000000f]: 0x34

我这里有我现在的片段

        string test = "ABCD1234";
        byte[] LogoDataBy = ASCIIEncoding.ASCII.GetBytes(test);
        var newArray = new byte[16];

        var difference = newArray.Length - LogoDataBy.Length;

        for (int i = 0; i < LogoDataBy.Length; i++)
        {
            newArray[difference + i] = LogoDataBy[i];
        }

有没有更有效的方法来做到这一点？

Answer 1

我建议从Array.Copy这样开始:

string test = "ABCD1234";
byte[] LogoDataBy = ASCIIEncoding.ASCII.GetBytes(test);
var newArray = new byte[16];

var startAt = newArray.Length - LogoDataBy.Length;
Array.Copy(LogoDataBy, 0, newArray, startAt, LogoDataBy.Length);

如果你真的需要速度,你也可以这样做Buffer.BlockCopy:

string test = "ABCD1234";
byte[] LogoDataBy = ASCIIEncoding.ASCII.GetBytes(test);
var newArray = new byte[16];

var startAt = newArray.Length - LogoDataBy.Length;
Buffer.BlockCopy(LogoDataBy, 0, newArray, startAt, LogoDataBy.Length);

请注意,我没有检查您提供的阵列的长度 - 您应该注意它足够大.

Answer 2

根据您如何定义“更高效”，这可能是值得做的：

var newArray =
    Enumerable
        .Repeat<Byte>(0, 16 - LogoDataBy.Length)
        .Concat(LogoDataBy)
        .ToArray();

这在计算上可能不是更有效，但在使代码清晰和可维护方面，您可能会认为这是一种有效的编码方式。