Is there a reason why It.Is only accepts an expression, thus breaking type inference?

[WillSullivan]

I seem to remember you could just pass in a regular old value and it would be satisfied

It.Is(1) 

rather than

It.Is<int>(x => 1 == x ) // ERRY GOTDAMN TIME

Seems like it's trivial to implement, like this replacement

public static class Is
{
    public static TValue Exactly<TValue>(TValue value)
    {
        return Match<TValue>.Create<TValue>(v => EqualityComparer<TValue>.Default.Equals(value, v));
    }
}

So why not? Is there some corner case or issue when processing this Match down the road? I can understand that there are assumptions about what is considered equality, but if we know what the deal is going in, why do we have to bash ourselves over the head with the Expression route?

[stakx]

@WillSullivan: Why would you choose to write It.Is(1) if you can just as well write 1?

[WillSullivan]

@stakx Forgot about this stinker right here. Of course that's the exact way it's done. At the time I wrote this I was trying to remember how I had previously been able to pass in a constant value. Was just looking in the wrong place. I'll clean up my mess, thanks.

[stakx]

Thanks for the response... no worries! ;-)