C# XML 反序列化 - 性能问题
本文关键字:性能 问题 反序列化 XML | 更新日期: 2023-09-27 17:56:13
我必须在反序列化之前解析XML文档。重用解析的结果(通过创建读取器)是否有意义,或者没有性能差异?
var root = XDocument.Parse(message).Root;
var type = mes3.Name.LocalName;
if (type == typeOf(SomeType))
{
var reader = root.CreateReader();
var serializer = new XmlSerializer(typeof(SomeType));
var someType = serializer.Deserialize(reader);
}
最好通过创建读取器来重用解析的结果,而不是直接使用字符串消息进行反序列化,因为在去浆化期间不会再次解析字符串。
如果我们使用快速而肮脏的演示程序:
using System;
using System.Text;
using System.Xml.Serialization;
using System.IO;
using System.Xml.Linq;
using System.Diagnostics;
namespace Sample_04_03_2014_01
{
public class Sample
{
public string Name { get; set; }
}
class Program
{
static void Main(string[] args)
{
Sample s = new Sample();
s.Name = "Hello";
var serializer = new XmlSerializer(typeof(Sample));
var sb = new StringBuilder();
using (var sw = new StringWriter(sb))
{
serializer.Serialize(sw, s);
}
string serialized = sb.ToString();
Console.WriteLine(serialized);
var root = XDocument.Parse(serialized).Root;
Sample someType1 = null;
Stopwatch stopWatch1 = new Stopwatch();
stopWatch1.Start();
for (int i = 0; i < 100000; i++)
{
var serializer1 = new XmlSerializer(typeof(Sample));
using (var reader = root.CreateReader())
{
someType1 = (Sample)serializer1.Deserialize(reader);
}
}
stopWatch1.Stop();
Console.WriteLine(someType1.Name);
Console.WriteLine(stopWatch1.Elapsed);
Sample someType2 = null;
Stopwatch stopWatch2 = new Stopwatch();
stopWatch2.Start();
for (int i = 0; i < 100000; i++)
{
var serializer2 = new XmlSerializer(typeof(Sample));
using (TextReader reader = new StringReader(serialized))
{
someType2 = (Sample)serializer2.Deserialize(reader);
}
}
stopWatch2.Stop();
Console.WriteLine(someType2.Name);
Console.WriteLine(stopWatch2.Elapsed);
}
}
}
然后,我们获得第一种方法(CreateReader)~30%的性能提升。
您好 00:00:00.8825465 您好 00:00:01.2636450
您的代码具有与以下代码相同的性能:
using (TextReader reader = new StringReader(message))
{
result = serializer.Deserialize(reader);
}
通过你有额外的对象.....
我用这个
using (FileStream xmlFile = new FileStream(filePath, FileMode.Open))
{
resultclass = (resultclasstype)serializer.Deserialize(xmlFile);
}