为什么Interlocked.Add没有超载接受双打作为参数？

Question

I fully appreciate the atomicity that the Threading.Interlocked class provides; I don't understand, though, why the Add function only offers two overloads: one for Integers, another for Longs. Why not Doubles, or any other numeric type for that matter?

Clearly, the intended method for changing a Double is CompareExchange; I am GUESSING this is because modifying a Double is a more complex operation than modifying an Integer. Still it isn't clear to me why, if CompareExchange and Add can both accept Integers, they can't also both accept Doubles.

Answer 1

Others have addressed the "why?". It is easy however to roll your own Add(ref double, double) , using the CompareExchange primitive:

public static double Add(ref double location1, double value)
{
    double newCurrentValue = location1; // non-volatile read, so may be stale
    while (true)
    {
        double currentValue = newCurrentValue;
        double newValue = currentValue + value;
        newCurrentValue = Interlocked.CompareExchange(ref location1, newValue, currentValue);
        if (newCurrentValue == currentValue)
            return newValue;
    }
}

CompareExchange sets the value of location1 to be newValue , if the current value equals currentValue . As it does so in an atomic, thread-safe way, we can rely on it alone without resorting to locks.

Why the while (true) loop? Loops like this are standard when implementing optimistically concurrent algorithms. CompareExchange will not change location1 if the current value is different from currentValue . I initialized currentValue to location1 - doing a non-volatile read (which might be stale, but that does not change the correctness, as CompareExchange will check the value). If the current value (still) is what we had read from location , CompareExchange will change the value to newValue . If not, we have to retry CompareExchange with the new current value, as returned by CompareExchange .

If the value is changed by another thread until the time of our next CompareExchange again, it will fail again, necessitating another retry - and this can in theory go on forever, hence the loop. Unless you are constantly changing the value from multiple threads, CompareExchange will most likely be called only once, if the current value is still what the non-volatile read of location1 yielded, or twice, if it was different.

Answer 2

The Interlocked class wraps around the Windows API Interlocked** functions.

These are, in turn, wrapping around the native processor API, using the LOCK instruction prefix for x86. It only supports prefixing the following instructions:

BT, BTS, BTR, BTC, XCHG, XADD, ADD, OR, ADC, SBB, AND, SUB, XOR, NOT, NEG, INC, DEC

You'll note that these, in turn, pretty much map to the interlocked methods. Unfortunately, the ADD functions for non-integer types are not supported here. Add for 64bit longs is supported on 64bit platforms.

Here's a great article discussing lock semantics on the instruction level .

Answer 3

As Reed Copsey has said, the Interlocked operations map (via Windows API functions) to instructions supported directly by the x86/x64 processors. Given that one of those functions is XCHG, you can do an atomic XCHG operation without really caring what the bits at the target location represent. In other words, the code can "pretend" that the 64-bit floating point number you are exchanging is in fact a 64-bit integer, and the XCHG instruction won't know the difference. Thus, .Net can provide Interlocked.Exchange functions for floats and doubles by "pretending" that they are integers and long integers, respectively.

However, all of the other operations actually do operate on the individual bits of the destination, and so they won't work unless the values actually represent integers (or bit arrays in some cases.)

Answer 4

I suspect that there are two reasons.

1. The processors targeted by .Net support interlocked increment only for integer types. I believe this is the LOCK prefix on x86, probably similar instructions exist for other processors.
2. Adding one to a floating point number can result in the same number if it is big enough, so I'm not sure if you can call that an increment. Perhaps the framework designers are trying to avoid nonintuitive behavior in this case.