如何在c#中获得具有最佳性能的线程安全计数器?
这很简单:
public static long GetNextValue()
{
long result;
lock (LOCK)
{
result = COUNTER++;
}
return result;
}
但是有没有更快的选择呢?
当前回答
如前所述，使用联锁。增量
来自MS的代码示例:
The following example determines how many random numbers that range from 0 to 1,000 are required to generate 1,000 random numbers with a midpoint value. To keep track of the number of midpoint values, a variable, midpointCount, is set equal to 0 and incremented each time the random number generator returns a midpoint value until it reaches 10,000. Because three threads generate the random numbers, the Increment(Int32) method is called to ensure that multiple threads don't update midpointCount concurrently. Note that a lock is also used to protect the random number generator, and that a CountdownEvent object is used to ensure that the Main method doesn't finish execution before the three threads.
using System;
using System.Threading;
public class Example
{
const int LOWERBOUND = 0;
const int UPPERBOUND = 1001;
static Object lockObj = new Object();
static Random rnd = new Random();
static CountdownEvent cte;
static int totalCount = 0;
static int totalMidpoint = 0;
static int midpointCount = 0;
public static void Main()
{
cte = new CountdownEvent(1);
// Start three threads.
for (int ctr = 0; ctr <= 2; ctr++) {
cte.AddCount();
Thread th = new Thread(GenerateNumbers);
th.Name = "Thread" + ctr.ToString();
th.Start();
}
cte.Signal();
cte.Wait();
Console.WriteLine();
Console.WriteLine("Total midpoint values: {0,10:N0} ({1:P3})",
totalMidpoint, totalMidpoint/((double)totalCount));
Console.WriteLine("Total number of values: {0,10:N0}",
totalCount);
}
private static void GenerateNumbers()
{
int midpoint = (UPPERBOUND - LOWERBOUND) / 2;
int value = 0;
int total = 0;
int midpt = 0;
do {
lock (lockObj) {
value = rnd.Next(LOWERBOUND, UPPERBOUND);
}
if (value == midpoint) {
Interlocked.Increment(ref midpointCount);
midpt++;
}
total++;
} while (midpointCount < 10000);
Interlocked.Add(ref totalCount, total);
Interlocked.Add(ref totalMidpoint, midpt);
string s = String.Format("Thread {0}:\n", Thread.CurrentThread.Name) +
String.Format(" Random Numbers: {0:N0}\n", total) +
String.Format(" Midpoint values: {0:N0} ({1:P3})", midpt,
((double) midpt)/total);
Console.WriteLine(s);
cte.Signal();
}
}
// The example displays output like the following:
// Thread Thread2:
// Random Numbers: 2,776,674
// Midpoint values: 2,773 (0.100 %)
// Thread Thread1:
// Random Numbers: 4,876,100
// Midpoint values: 4,873 (0.100 %)
// Thread Thread0:
// Random Numbers: 2,312,310
// Midpoint values: 2,354 (0.102 %)
//
// Total midpoint values: 10,000 (0.100 %)
// Total number of values: 9,965,084
下面的示例与前面的示例类似，只是它使用Task类而不是线程过程来生成50,000个随机中点整数。在本例中，lambda表达式替换了GenerateNumbers线程过程和Task调用。方法消除了对CountdownEvent对象的需要。
using System;
using System.Collections.Generic;
using System.Threading;
using System.Threading.Tasks;
public class Example
{
const int LOWERBOUND = 0;
const int UPPERBOUND = 1001;
static Object lockObj = new Object();
static Random rnd = new Random();
static int totalCount = 0;
static int totalMidpoint = 0;
static int midpointCount = 0;
public static void Main()
{
List<Task> tasks = new List<Task>();
// Start three tasks.
for (int ctr = 0; ctr <= 2; ctr++)
tasks.Add(Task.Run( () => { int midpoint = (UPPERBOUND - LOWERBOUND) / 2;
int value = 0;
int total = 0;
int midpt = 0;
do {
lock (lockObj) {
value = rnd.Next(LOWERBOUND, UPPERBOUND);
}
if (value == midpoint) {
Interlocked.Increment(ref midpointCount);
midpt++;
}
total++;
} while (midpointCount < 50000);
Interlocked.Add(ref totalCount, total);
Interlocked.Add(ref totalMidpoint, midpt);
string s = String.Format("Task {0}:\n", Task.CurrentId) +
String.Format(" Random Numbers: {0:N0}\n", total) +
String.Format(" Midpoint values: {0:N0} ({1:P3})", midpt,
((double) midpt)/total);
Console.WriteLine(s); } ));
Task.WaitAll(tasks.ToArray());
Console.WriteLine();
Console.WriteLine("Total midpoint values: {0,10:N0} ({1:P3})",
totalMidpoint, totalMidpoint/((double)totalCount));
Console.WriteLine("Total number of values: {0,10:N0}",
totalCount);
}
}
// The example displays output like the following:
// Task 3:
// Random Numbers: 10,855,250
// Midpoint values: 10,823 (0.100 %)
// Task 1:
// Random Numbers: 15,243,703
// Midpoint values: 15,110 (0.099 %)
// Task 2:
// Random Numbers: 24,107,425
// Midpoint values: 24,067 (0.100 %)
//
// Total midpoint values: 50,000 (0.100 %)
// Total number of values: 50,206,378
https://learn.microsoft.com/en-us/dotnet/api/system.threading.interlocked.increment?view=netcore-3.0
其他回答
试试Interlocked。增量
这样会更简单:
return Interlocked.Increment(ref COUNTER);
MSDN联锁。增量
我建议你在系统中使用。net内置的联锁增量。线程库。
下面的代码将通过引用增加一个长变量，并且是完全线程安全的:
Interlocked.Increment(ref myNum);
来源:http://msdn.microsoft.com/en-us/library/dd78zt0c.aspx
正如其他人推荐的那样，联锁。Increment将比lock()具有更好的性能。只要看一下IL和Assembly，你就会看到Increment变成了一个“总线锁”语句，它的变量直接被增量(x86)或“添加”到(x64)。
这个“总线锁”语句锁定总线，以防止另一个CPU在调用CPU执行操作时访问总线。现在，看一下c# lock()语句的IL。在这里，您将看到为了开始或结束一个节而调用Monitor。
换句话说，. net lock()语句比. net Interlocked.Increment要做的多得多。
如果你想做的只是增加一个变量，Interlock。增量会更快。回顾所有Interlocked方法，以查看可用的各种原子操作，并找到适合您需要的操作。当您想要做更复杂的事情时，比如多个相互关联的递增/递减，或者序列化对比整数更复杂的资源的访问时，可以使用lock()。
如前所述，使用联锁。增量
来自MS的代码示例:
The following example determines how many random numbers that range from 0 to 1,000 are required to generate 1,000 random numbers with a midpoint value. To keep track of the number of midpoint values, a variable, midpointCount, is set equal to 0 and incremented each time the random number generator returns a midpoint value until it reaches 10,000. Because three threads generate the random numbers, the Increment(Int32) method is called to ensure that multiple threads don't update midpointCount concurrently. Note that a lock is also used to protect the random number generator, and that a CountdownEvent object is used to ensure that the Main method doesn't finish execution before the three threads.
using System;
using System.Threading;
public class Example
{
const int LOWERBOUND = 0;
const int UPPERBOUND = 1001;
static Object lockObj = new Object();
static Random rnd = new Random();
static CountdownEvent cte;
static int totalCount = 0;
static int totalMidpoint = 0;
static int midpointCount = 0;
public static void Main()
{
cte = new CountdownEvent(1);
// Start three threads.
for (int ctr = 0; ctr <= 2; ctr++) {
cte.AddCount();
Thread th = new Thread(GenerateNumbers);
th.Name = "Thread" + ctr.ToString();
th.Start();
}
cte.Signal();
cte.Wait();
Console.WriteLine();
Console.WriteLine("Total midpoint values: {0,10:N0} ({1:P3})",
totalMidpoint, totalMidpoint/((double)totalCount));
Console.WriteLine("Total number of values: {0,10:N0}",
totalCount);
}
private static void GenerateNumbers()
{
int midpoint = (UPPERBOUND - LOWERBOUND) / 2;
int value = 0;
int total = 0;
int midpt = 0;
do {
lock (lockObj) {
value = rnd.Next(LOWERBOUND, UPPERBOUND);
}
if (value == midpoint) {
Interlocked.Increment(ref midpointCount);
midpt++;
}
total++;
} while (midpointCount < 10000);
Interlocked.Add(ref totalCount, total);
Interlocked.Add(ref totalMidpoint, midpt);
string s = String.Format("Thread {0}:\n", Thread.CurrentThread.Name) +
String.Format(" Random Numbers: {0:N0}\n", total) +
String.Format(" Midpoint values: {0:N0} ({1:P3})", midpt,
((double) midpt)/total);
Console.WriteLine(s);
cte.Signal();
}
}
// The example displays output like the following:
// Thread Thread2:
// Random Numbers: 2,776,674
// Midpoint values: 2,773 (0.100 %)
// Thread Thread1:
// Random Numbers: 4,876,100
// Midpoint values: 4,873 (0.100 %)
// Thread Thread0:
// Random Numbers: 2,312,310
// Midpoint values: 2,354 (0.102 %)
//
// Total midpoint values: 10,000 (0.100 %)
// Total number of values: 9,965,084
下面的示例与前面的示例类似，只是它使用Task类而不是线程过程来生成50,000个随机中点整数。在本例中，lambda表达式替换了GenerateNumbers线程过程和Task调用。方法消除了对CountdownEvent对象的需要。
using System;
using System.Collections.Generic;
using System.Threading;
using System.Threading.Tasks;
public class Example
{
const int LOWERBOUND = 0;
const int UPPERBOUND = 1001;
static Object lockObj = new Object();
static Random rnd = new Random();
static int totalCount = 0;
static int totalMidpoint = 0;
static int midpointCount = 0;
public static void Main()
{
List<Task> tasks = new List<Task>();
// Start three tasks.
for (int ctr = 0; ctr <= 2; ctr++)
tasks.Add(Task.Run( () => { int midpoint = (UPPERBOUND - LOWERBOUND) / 2;
int value = 0;
int total = 0;
int midpt = 0;
do {
lock (lockObj) {
value = rnd.Next(LOWERBOUND, UPPERBOUND);
}
if (value == midpoint) {
Interlocked.Increment(ref midpointCount);
midpt++;
}
total++;
} while (midpointCount < 50000);
Interlocked.Add(ref totalCount, total);
Interlocked.Add(ref totalMidpoint, midpt);
string s = String.Format("Task {0}:\n", Task.CurrentId) +
String.Format(" Random Numbers: {0:N0}\n", total) +
String.Format(" Midpoint values: {0:N0} ({1:P3})", midpt,
((double) midpt)/total);
Console.WriteLine(s); } ));
Task.WaitAll(tasks.ToArray());
Console.WriteLine();
Console.WriteLine("Total midpoint values: {0,10:N0} ({1:P3})",
totalMidpoint, totalMidpoint/((double)totalCount));
Console.WriteLine("Total number of values: {0,10:N0}",
totalCount);
}
}
// The example displays output like the following:
// Task 3:
// Random Numbers: 10,855,250
// Midpoint values: 10,823 (0.100 %)
// Task 1:
// Random Numbers: 15,243,703
// Midpoint values: 15,110 (0.099 %)
// Task 2:
// Random Numbers: 24,107,425
// Midpoint values: 24,067 (0.100 %)
//
// Total midpoint values: 50,000 (0.100 %)
// Total number of values: 50,206,378
https://learn.microsoft.com/en-us/dotnet/api/system.threading.interlocked.increment?view=netcore-3.0
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