0

如何以最短和有效的方法将各种字符串连接成一个字符串。目前我正在按照以下方式进行

string str1 = string.Format(BitConverter
       .ToString(pass_packet, local_index + 511, 1) +
       BitConverter.ToString(pass_packet, local_index + 510, 1) +
       BitConverter.ToString(pass_packet, local_index + 509, 1) +
       BitConverter.ToString(pass_packet, local_index + 508, 1) + 
       ... + BitConverter.ToString(pass_packet, local_index + 400, 1));
4

4 回答 4

5

鉴于测试数据，我认为您根本不需要连接。获取字节数组，反转它们，然后通过单个 BitConverter.ToString 调用！

class Program
{
    static int ITERATIONS = 100000;
    static void Main(string[] args)
    {
        var pass_packet = Enumerable.Range(0, 1024).Select(i => (byte)i).ToArray();

        int local_index = 5;

        var sw = Stopwatch.StartNew();
        var result = StringBuilderTEST(pass_packet, local_index);

        Console.WriteLine(result + " in {0}ms", sw.ElapsedMilliseconds);

        //second option
        sw.Restart();
        var result2 = ArrayReversalTEST(pass_packet, local_index);
        Console.WriteLine(result2 + " in {0}ms", sw.ElapsedMilliseconds);

        sw.Restart();
        var result3 = ArrayReversal2TEST(pass_packet, local_index);
        Console.WriteLine(result3 + " in {0}ms", sw.ElapsedMilliseconds);

        sw.Restart();
        var result4 = StupidlyFastTEST(pass_packet, local_index);
        Console.WriteLine(result4 + " in {0}ms", sw.ElapsedMilliseconds);

        Console.WriteLine("Results are equal?  " + (result == result2 && result == result3 && result == result4));
        Console.ReadLine();
    }

    private static string StringBuilderTEST(byte[] pass_packet, int local_index)
    {
        string result = null;
        for (int b = 0; b < ITERATIONS; b++)
        {
            var sb = new StringBuilder();
            for (int i = 511; i >= 400; i--)
                sb.Append(BitConverter.ToString(pass_packet, local_index + i, 1));
            result = sb.ToString();
        }
        return result;
    }

    private static string ArrayReversalTEST(byte[] pass_packet, int local_index)
    {
        string result = null;
        for (int b = 0; b < ITERATIONS; b++)
        {
            var selectedData = pass_packet.Skip(400 + local_index).Take(112).Reverse().ToArray();
            result = BitConverter.ToString(selectedData).Replace("-", "");
        }
        return result;
    }

    private static string ArrayReversal2TEST(byte[] pass_packet, int local_index)
    {
        string result = null;
        for (int b = 0; b < ITERATIONS; b++)
        {
            var tempArray = new byte[112];
            Array.Copy(pass_packet, 400 + local_index, tempArray, 0, 112);
            Array.Reverse(tempArray);
            result = BitConverter.ToString(tempArray).Replace("-", "");
        }
        return result;
    }

    private static string StupidlyFastTEST(byte[] pass_packet, int local_index)
    {
        string result = null;
        string hex = "0123456789ABCDEF";
        for (int it = 0; it < ITERATIONS; it++)
        {
            var tempArray = new char[112 * 2];
            int tempArrayIndex = 0;
            for (int i = 511; i >= 400; i--)
            {
                var b = pass_packet[local_index + i];
                tempArray[tempArrayIndex++] = hex[b >> 4];
                tempArray[tempArrayIndex++] = hex[b & 0x0F];
            }
            result = new string(tempArray);
        }
        return result;
    }
}

结果：

Test 1 in 478ms
Test 2 in 1134ms
Test 3 in 516ms
Test 4 in 114ms
Results are equal?  True

正如您所看到的，我重写代码的前两次尝试效率不高——尤其是考虑到创建和维护所需的额外时间。然而，一些快速测试表明，这是由于使结果相同所需的 String.Replace - 因为默认转换器在数组中的每个字节对之间放置“-”，由于单字节长度，原始算法没有看到.

//Without String.Replace in tests 2 and 3
Test 1 in 475ms
Test 2 in 704ms
Test 3 in 92ms
Test 4 in 115ms
Results are equal?  False

正如您在原始性能方面所见，Test3 是最快的 - 尽管默认输出在每个字节之间包含“-”。

Test4 用快速手动版本替换了转换器 - 消除了字节分隔符 - 并且是原始结果最快的。我怀疑缓存临时数组，并用更大的 256 元素十六进制数组替换除法和模数数学* 会大大加快速度，但鉴于这一点已被证明会停在这里。

*编辑，用位运算替换除法和模数以显着加快速度。

于 2013-08-21T09:11:29.347 回答
2

我会这样做：

StringBuilder sb = new StringBuilder();

for (int i = 511; i >= 400; --i)
    sb.Append(BitConverter.ToString(pass_packet, local_index + i, 1));

string str1 = sb.ToString();
于 2013-08-21T09:09:36.390 回答
0

是你的string.Concat朋友。

string.Concat(BitConverter.ToString(pass_packet, local_index + 511, 1),
              BitConverter.ToString(pass_packet, local_index + 510, 1), 
              BitConverter.ToString(pass_packet, local_index + 509, 1),
              BitConverter.ToString(pass_packet, local_index + 508, 1),                  
              BitConverter.ToString(pass_packet, local_index + 400, 1));

有两个有趣的重载：string.Concat(params Object[] args)和 string.Concat(params string[] values)。两者都接受可变长度的参数（正如我所做的那样）或正确类型的数组。甚至有接受IEnumerable<string>或IEnumerable<T>其中T任何一种类型的重载。

于 2013-08-21T08:57:47.200 回答
0

这有点棘手，但我认为这可以完成工作。

string myString = new String(Enumerable.Range(400,111).SelectMany(x => BitConverter.ToString(pass_packet, x + 509, 1)).ToArray());
于 2013-08-21T09:02:31.150 回答