我想制作一个采用二进制字符串并将其转换为十进制字符串的程序。 我知道你是怎么做到的:
Convert.ToString(00001000, 2);
要将单个二进制数字转换为单个十进制数字,但是我如何在字符串中获取一堆二进制数字,像这样:
00001000 00000101 00001100 00001100 00001111
00010111 00001111 00010010 00001100 00000100
转换为十进制数字字符串,如下所示:
8 5 12 12 15
23 15 18 12 4
感谢您的帮助。
答案 0 :(得分:3)
将指定基数中的数字的字符串表示形式转换为 等效的32位带符号整数。
int output = Convert.ToInt32(input, 2);
如果它在用空格隔开的列表中,只需使用string.Split
返回一个字符串数组,该数组包含此实例中的子字符串 由指定的字符串或Unicode元素分隔的 字符数组。
var input = "00001000 00000101 00001100 00001100 00001111";
var results = input.Split(' ')
.Select(x => Convert.ToInt32(x, 2))
.ToList();
Console.WriteLine(string.Join(",",results))
输出
8,5,12,12,15
对于那些在家里玩的人来说,这是将二进制字符串转换为int
的一种毫无意义的方法,而完全忽略了字节序
注意:比例尺只是预期的项目数,可以轻松地用List.Add()替换它,而且成本低廉
[Test("BitShift", "", false)]
public unsafe List<int> Convert(string input, int scale)
{
var list = new int[scale];
fixed (int* plist = list)
fixed (char* pInput = input)
{
var pLen = pInput + input.Length;
var val = 0;
var pl = plist;
for (var p = pInput; p < pLen; p++)
if (*p != ' ')
val = (val << 1) + (*p - 48);
else
{
*pl = val;
pl++;
val = 0;
}
*pl = val;
}
return list.ToList();
}
以下是您享受娱乐的一些基准
----------------------------------------------------------------------------
Mode : Release (64Bit)
Test Framework : .NET Framework 4.7.1 (CLR 4.0.30319.42000)
----------------------------------------------------------------------------
Operating System : Microsoft Windows 10 Pro
Version : 10.0.17134
----------------------------------------------------------------------------
CPU Name : Intel(R) Core(TM) i7-3770K CPU @ 3.50GHz
Description : Intel64 Family 6 Model 58 Stepping 9
Cores (Threads) : 4 (8) : Architecture : x64
Clock Speed : 3901 MHz : Bus Speed : 100 MHz
L2Cache : 1 MB : L3Cache : 8 MB
----------------------------------------------------------------------------
测试1
--- Standard input ------------------------------------------------------------
| Value | Average | Fastest | Cycles | Garbage | Test | Gain |
--- Scale 10 --------------------------------------------------- Time 0.654 ---
| BitShift | 2.433 µs | 2.045 µs | 12.379 K | 0.000 B | Pass | 0.00 % |
| Convert | 7.348 µs | 6.137 µs | 29.792 K | 0.000 B | Pass | -202.08 % |
--- Scale 100 -------------------------------------------------- Time 0.618 ---
| BitShift | 6.629 µs | 6.137 µs | 26.656 K | 0.000 B | Pass | 0.00 % |
| Convert | 37.805 µs | 33.320 µs | 136.246 K | 0.000 B | Pass | -470.31 % |
--- Scale 1,000 ------------------------------------------------ Time 0.978 ---
| BitShift | 56.409 µs | 48.519 µs | 201.670 K | 0.000 B | Pass | 0.00 % |
| Convert | 335.717 µs | 286.730 µs | 1.181 M | 0.000 B | Pass | -495.15 % |
--- Scale 10,000 ----------------------------------------------- Time 3.723 ---
| BitShift | 488.767 µs | 415.919 µs | 1.715 M | 0.000 B | Pass | 0.00 % |
| Convert | 3.209 ms | 2.720 ms | 10.955 M | 0.000 B | Pass | -556.46 % |
--- Scale 100,000 --------------------------------------------- Time 39.832 ---
| BitShift | 5.110 ms | 4.195 ms | 17.788 M | 0.000 B | Pass | 0.00 % |
| Convert | 42.987 ms | 35.171 ms | 141.218 M | 0.000 B | Pass | -741.28 % |
-------------------------------------------------------------------------------
答案 1 :(得分:2)
要转换成这样的单个
:Convert.ToInt32("00001000", 2)
我认为我们可以使用Regex替换所有二进制数字:
Regex.Replace(binaryNumbers, @"\b\d+\b", new MatchEvaluator((m) => {
return Convert.ToInt32(m.Value, 2).ToString();
}));