使用MemoryStream和CryptoStream时清除Gen2和LOH中的堆

问题描述

我正在尝试进行一些测量,以了解如果添加了“加密/解密过程”,则会占用多少空间。另外,我正在比较使用FileStream或返回MemoryStream(在某些情况下需要)的不同方法

看起来像大文件一样保留在内存中(Gen2和LOH)。如何完全清除堆(我想在FileStream方法中看到相同的Gen2结果)?

我正在使用using关键字。但是看起来没有希望了!我还减小了认的“缓冲区大小”,如下面的代码所示。但是第二代中我仍然有数字

BenchmarkDotNet=v0.12.1,OS=Windows 10.0.19041.572 (2004/?/20H1)
Intel Core i9-10920X cpu 3.50GHz,1 cpu,24 logical and 12 physical cores
  [Host]     : .NET Framework 4.8 (4.8.4250.0),X86 LegacyJIT
  DefaultJob : .NET Framework 4.8 (4.8.4250.0),X86 LegacyJIT

文件流结果

|              Method |      Mean |     Error |    StdDev |     Gen 0 |    Gen 1 | Gen 2 |   Allocated |
|-------------------- |----------:|----------:|----------:|----------:|---------:|------:|------------:|
| TXT300BYTES_Decrypt |  2.500 ms | 0.0444 ms | 0.0593 ms |   19.5313 |        - |     - |   105.11 KB |
|    PDF500KB_Decrypt | 12.909 ms | 0.2561 ms | 0.4348 ms |  187.5000 |  15.6250 |     - |  1019.59 KB |
|      PDF1MB_Decrypt | 14.125 ms | 0.2790 ms | 0.4001 ms |  406.2500 |  15.6250 |     - |  2149.96 KB |
|     TIFF1MB_Decrypt | 10.087 ms | 0.1949 ms | 0.1728 ms |  437.5000 |  31.2500 |     - |  2329.37 KB |
|     TIFF5MB_Decrypt | 22.779 ms | 0.4316 ms | 0.4239 ms | 2000.0000 | 187.5000 |     - | 10434.34 KB |
|    TIFF10MB_Decrypt | 38.467 ms | 0.7382 ms | 0.8205 ms | 3857.1429 | 285.7143 |     - | 20144.01 KB |

内存流结果

|              Method |      Mean |     Error |    StdDev |     Gen 0 |     Gen 1 |    Gen 2 |   Allocated |
|-------------------- |----------:|----------:|----------:|----------:|----------:|---------:|------------:|
| TXT300BYTES_Decrypt |  1.673 ms | 0.0098 ms | 0.0092 ms |   27.3438 |    1.9531 |        - |   147.69 KB |
|    PDF500KB_Decrypt |  9.956 ms | 0.1407 ms | 0.1248 ms |  328.1250 |  328.1250 | 328.1250 |  2316.08 KB |
|      PDF1MB_Decrypt | 11.998 ms | 0.0622 ms | 0.0486 ms |  921.8750 |  546.8750 | 531.2500 |   4737.8 KB |
|     TIFF1MB_Decrypt |  9.252 ms | 0.0973 ms | 0.0910 ms |  953.1250 |  671.8750 | 500.0000 |  4902.34 KB |
|     TIFF5MB_Decrypt | 24.220 ms | 0.1105 ms | 0.0980 ms | 2531.2500 |  718.7500 | 468.7500 | 20697.43 KB |
|    TIFF10MB_Decrypt | 41.463 ms | 0.5678 ms | 0.5033 ms | 4833.3333 | 1500.0000 | 916.6667 | 40696.31 KB |
public static class Constants
{
    public const int BufferSize = 40960; // Default is  81920
}

文件解密方法

public class DescryptionService
{
    public async Task<string> DecryptFileAsync(string sourcePath)
    {
        var tempFilePath = SecurityFileHelper.CreateTempFile();
        using var sourceStream = new FileStream(sourcePath,FileMode.Open,FileAccess.Read,FileShare.Read);
        var keyBytes = Convert.FromBase64String(_key);
        using var destinationStream = new FileStream(tempFilePath,FileAccess.ReadWrite,FileShare.ReadWrite);
        using var provider = new AesCryptoServiceProvider();

        var IV = new byte[provider.IV.Length];
        await sourceStream.ReadAsync(IV,IV.Length);

        using var cryptoTransform = provider.CreateDecryptor(keyBytes,IV);
        using var cryptoStream = new CryptoStream(sourceStream,cryptoTransform,CryptoStreamMode.Read);
        await cryptoStream.copyToAsync(destinationStream,Constants.BufferSize);

        return tempFilePath;
    }
}

内存解密方法

public class DescryptionService
{
    public async Task<Stream> DecryptStreamAsync(Stream sourceStream)
    {
        var memoryStream = new MemoryStream();

        if (sourceStream.Position != 0) sourceStream.Position = 0;
        var tempFilePath = SecurityFileHelper.CreateTempFile();
        try
        {
            var keyBytes = Convert.FromBase64String(_key);
            using var destinationStream = new FileStream(tempFilePath,FileShare.ReadWrite);
            using var provider = new AesCryptoServiceProvider();

            var IV = new byte[provider.IV.Length];
            await sourceStream.ReadAsync(IV,IV.Length);

            using var cryptoTransform = provider.CreateDecryptor(keyBytes,IV);
            using var cryptoStream = new CryptoStream(sourceStream,CryptoStreamMode.Read);
            await cryptoStream.copyToAsync(destinationStream,Constants.BufferSize);
            destinationStream.Position = 0;
            await destinationStream.copyToAsync(memoryStream,Constants.BufferSize);
            await memoryStream.FlushAsync();
            memoryStream.Position = 0;
        }
        finally
        {
            if (File.Exists(tempFilePath))
                File.Delete(tempFilePath);
        }
        return memoryStream;
    }
}
// Calling it like this
using var encryptedStream = File.OpenRead("some file path");
var svc = new DecryptionService();
using var decryptedStream = await svc.DecryptStreamAsync(encryptedStream);

通过这种方式,我还添加了以下几行:

decryptedStream.Position = 0;
decryptedStream.SetLength(0);
decryptedStream.Capacity = 0; // <<< this one will null bytes in memory stream

仍然有这些结果

|              Method |      Mean |     Error |    StdDev |    Median |     Gen 0 |     Gen 1 |    Gen 2 |   Allocated |
|-------------------- |----------:|----------:|----------:|----------:|----------:|----------:|---------:|------------:|
| TXT300BYTES_Decrypt |  1.659 ms | 0.0322 ms | 0.0301 ms |  1.662 ms |   27.3438 |    1.9531 |        - |   148.03 KB |
|    PDF500KB_Decrypt | 11.085 ms | 0.2829 ms | 0.8297 ms | 10.769 ms |  328.1250 |  328.1250 | 328.1250 |  2312.33 KB |
|      PDF1MB_Decrypt | 12.479 ms | 0.2029 ms | 0.3859 ms | 12.402 ms |  906.2500 |  562.5000 | 531.2500 |  4734.61 KB |
|     TIFF1MB_Decrypt |  9.352 ms | 0.0971 ms | 0.0861 ms |  9.359 ms |  953.1250 |  593.7500 | 500.0000 |     4908 KB |
|     TIFF5MB_Decrypt | 24.760 ms | 0.4752 ms | 0.4213 ms | 24.607 ms | 2593.7500 |  843.7500 | 531.2500 | 20715.76 KB |
|    TIFF10MB_Decrypt | 41.976 ms | 0.6657 ms | 0.5901 ms | 42.011 ms | 4833.3333 | 1500.0000 | 916.6667 | 40744.43 KB |

我想念什么?! :(

解决方法

我正在比较使用FileStream或返回MemoryStream之类的不同方法

看起来像大文件一样保留在内存中(Gen2和LOH)。如何完全清除堆(我想在FileStream方法中看到相同的Gen2结果)?

我不确定我是否了解清除堆的含义以及为什么要查看FileStream的Gen 2集合。

您正在使用的线束(BenchmarkDotNet)在two full memory cleanups之后强制执行every benchmark iteration。它确保每次基准测试迭代都以“干净堆”开始。为了确保GC的自调整性质(或其他任何类似内存泄漏的问题)不会影响其他基准,每个基准都在独立的进程中执行。此外,每1k次操作(基准调用)按比例缩放的集合数。这样可以对GC指标进行逐个比较。

您正在比较两种不同的方法,并且很可能(这是一个需要使用内存分析器进行验证的假设),其中一种方法分配了大对象,因此您获得了Gen 2集合。另一个没有。 它是给定解决方案的性能特征,您在实施业务逻辑时应该考虑它。例如:如果您的服务被认为是低延迟的,并且您不允许由Gen 2集合引起的长时间GC暂停,则应该选择不分配大对象的方法。

如果您想摆脱第2代集合,可以尝试使用以下方法池化内存: