我知道c#中实例化的值类型数组会自动填充该类型的默认值(例如bool为false, int为0，等等)。
是否有一种方法来自动填充一个不是默认的种子值的数组?无论是在创建或之后的内置方法(如Java的Arrays.fill())?假设我想要一个默认为true的布尔数组，而不是false。是否有一个内置的方法来做到这一点，或者你只需要通过一个for循环迭代数组?
// Example pseudo-code:
bool[] abValues = new[1000000];
Array.Populate(abValues, true);
// Currently how I'm handling this:
bool[] abValues = new[1000000];
for (int i = 0; i < 1000000; i++)
{
abValues[i] = true;
}
必须遍历数组并将每个值“重置”为true似乎效率不高。还有其他方法吗?也许通过翻转所有值?
在输入这个问题并思考之后，我猜默认值只是c#在幕后处理这些对象的内存分配的结果，所以我想这可能是不可能的。但我还是想确定一下!
并行实现怎么样
public static void InitializeArray<T>(T[] array, T value)
{
var cores = Environment.ProcessorCount;
ArraySegment<T>[] segments = new ArraySegment<T>[cores];
var step = array.Length / cores;
for (int i = 0; i < cores; i++)
{
segments[i] = new ArraySegment<T>(array, i * step, step);
}
var remaining = array.Length % cores;
if (remaining != 0)
{
var lastIndex = segments.Length - 1;
segments[lastIndex] = new ArraySegment<T>(array, lastIndex * step, array.Length - (lastIndex * step));
}
var initializers = new Task[cores];
for (int i = 0; i < cores; i++)
{
var index = i;
var t = new Task(() =>
{
var s = segments[index];
for (int j = 0; j < s.Count; j++)
{
array[j + s.Offset] = value;
}
});
initializers[i] = t;
t.Start();
}
Task.WaitAll(initializers);
}
当只初始化一个数组时，这段代码的力量是看不出来的，但我认为你绝对应该忘记“纯”for。
只是一个基准:
BenchmarkDotNet=v0.12.1, OS=Windows 10.0.18363.997 (1909/November2018Update/19H2)
Intel Core i7-6700HQ CPU 2.60GHz (Skylake), 1 CPU, 8 logical and 4 physical cores
.NET Core SDK=3.1.302
[Host] : .NET Core 3.1.6 (CoreCLR 4.700.20.26901, CoreFX 4.700.20.31603), X64 RyuJIT
.NET Core 3.1 : .NET Core 3.1.6 (CoreCLR 4.700.20.26901, CoreFX 4.700.20.31603), X64 RyuJIT
Job=.NET Core 3.1 Runtime=.NET Core 3.1
| Method | Mean | Error | StdDev |
|----------------- |---------:|----------:|----------:|
| EnumerableRepeat | 2.311 us | 0.0228 us | 0.0213 us |
| NewArrayForEach | 2.007 us | 0.0392 us | 0.0348 us |
| ArrayFill | 2.426 us | 0.0103 us | 0.0092 us |
[SimpleJob(BenchmarkDotNet.Jobs.RuntimeMoniker.NetCoreApp31)]
public class InitializeArrayBenchmark {
const int ArrayLength = 1600;
[Benchmark]
public double[] EnumerableRepeat() {
return Enumerable.Repeat(double.PositiveInfinity, ArrayLength).ToArray();
}
[Benchmark]
public double[] NewArrayForEach() {
var array = new double[ArrayLength];
for (int i = 0; i < array.Length; i++) {
array[i] = double.PositiveInfinity;
}
return array;
}
[Benchmark]
public double[] ArrayFill() {
var array = new double[ArrayLength];
Array.Fill(array, double.PositiveInfinity);
return array;
}
}
并行实现怎么样
public static void InitializeArray<T>(T[] array, T value)
{
var cores = Environment.ProcessorCount;
ArraySegment<T>[] segments = new ArraySegment<T>[cores];
var step = array.Length / cores;
for (int i = 0; i < cores; i++)
{
segments[i] = new ArraySegment<T>(array, i * step, step);
}
var remaining = array.Length % cores;
if (remaining != 0)
{
var lastIndex = segments.Length - 1;
segments[lastIndex] = new ArraySegment<T>(array, lastIndex * step, array.Length - (lastIndex * step));
}
var initializers = new Task[cores];
for (int i = 0; i < cores; i++)
{
var index = i;
var t = new Task(() =>
{
var s = segments[index];
for (int j = 0; j < s.Count; j++)
{
array[j + s.Offset] = value;
}
});
initializers[i] = t;
t.Start();
}
Task.WaitAll(initializers);
}
当只初始化一个数组时，这段代码的力量是看不出来的，但我认为你绝对应该忘记“纯”for。
我有点惊讶没有人做了非常简单，但超快的SIMD版本:
public static void PopulateSimd<T>(T[] array, T value) where T : struct
{
var vector = new Vector<T>(value);
var i = 0;
var s = Vector<T>.Count;
var l = array.Length & ~(s-1);
for (; i < l; i += s) vector.CopyTo(array, i);
for (; i < array.Length; i++) array[i] = value;
}
基准测试:(数据来自于Framework 4.8，但Core3.1在统计上是相同的)
| Method | N | Mean | Error | StdDev | Ratio | RatioSD |
|----------- |-------- |---------------:|---------------:|--------------:|------:|--------:|
| DarthGizka | 10 | 25.975 ns | 1.2430 ns | 0.1924 ns | 1.00 | 0.00 |
| Simd | 10 | 3.438 ns | 0.4427 ns | 0.0685 ns | 0.13 | 0.00 |
| | | | | | | |
| DarthGizka | 100 | 81.155 ns | 3.8287 ns | 0.2099 ns | 1.00 | 0.00 |
| Simd | 100 | 12.178 ns | 0.4547 ns | 0.0704 ns | 0.15 | 0.00 |
| | | | | | | |
| DarthGizka | 1000 | 201.138 ns | 8.9769 ns | 1.3892 ns | 1.00 | 0.00 |
| Simd | 1000 | 100.397 ns | 4.0965 ns | 0.6339 ns | 0.50 | 0.00 |
| | | | | | | |
| DarthGizka | 10000 | 1,292.660 ns | 38.4965 ns | 5.9574 ns | 1.00 | 0.00 |
| Simd | 10000 | 1,272.819 ns | 68.5148 ns | 10.6027 ns | 0.98 | 0.01 |
| | | | | | | |
| DarthGizka | 100000 | 16,156.106 ns | 366.1133 ns | 56.6564 ns | 1.00 | 0.00 |
| Simd | 100000 | 17,627.879 ns | 1,589.7423 ns | 246.0144 ns | 1.09 | 0.02 |
| | | | | | | |
| DarthGizka | 1000000 | 176,625.870 ns | 32,235.9957 ns | 1,766.9637 ns | 1.00 | 0.00 |
| Simd | 1000000 | 186,812.920 ns | 18,069.1517 ns | 2,796.2212 ns | 1.07 | 0.01 |
可以看到，在小于10000个元素时速度要快得多，超过10000个元素时速度仅略慢。
