问题
If many threads are calling GetNextNumber
simultaneously with the following code, GetNextNumber
will return 1 more times than any other numbers.
private class RoundRobbinNumber
{
private int _maxNumbers = 10;
private int _lastNumber;
private RoundRobbinNumber(int maxNumbers)
{
_maxNumbers = maxNumbers;
}
public int GetNextNumber()
{
int nextNumber = Interlocked.Increment(ref _lastNumber);
if (_lastNumber > _maxNumbers)
{
Interlocked.CompareExchange(ref _lastNumber, 1, _maxNumbers);
nextNumber = 1;
}
return nextNumber;
}
}
Is there a way to reset the _lastNumber
back to one, and reliably return an incremented number for each thread calling GetNextNumber()
, without having to use a lock?
回答1:
The trick is to do the operation in a loop until it is successful. I provide a general template for this approach in my answer here.
public int GetNextNumber()
{
int initial, computed;
do
{
initial = _lastNumber;
computed = initial + 1;
computed = computed > _maxNumbers ? computed = 1 : computed;
}
while (Interlocked.CompareExchange(ref _lastNumber, computed, initial) != initial);
return computed;
}
回答2:
Not sure if if helps anyone but this could be even simpler:
class RoundRobinNumber
{
private int _maxNumbers = 10;
private int _lastNumber = 0;
public RoundRobinNumber(int maxNumbers)
{
_maxNumbers = maxNumbers;
}
public int GetNextNumber()
{
int nextNumber = Interlocked.Increment(ref _lastNumber);
int result = nextNumber % _maxNumbers;
return result >= 0 ? result : -result;
}
}
回答3:
Andrey's answer without conditional statements:
using System;
namespace Utils
{
public class RoundRobinCounter
{
private int _max;
private int _currentNumber = 0;
public RoundRobinCounter(int max)
{
_max = max;
}
public int GetNext()
{
uint nextNumber = unchecked((uint)System.Threading.Interlocked.Increment(ref _currentNumber));
int result = (int)(nextNumber % _max);
return result;
}
}
}
And here is a .net fiddle running this code.
来源:https://stackoverflow.com/questions/10034289/how-should-i-increment-a-number-for-a-round-robin-threading-scenario-with-least