Q:

How should one define the sample variance for scalar input?

I was horrified to find recently that Matlab returns $0$ for the sample variance of a scalar input:
>> var(randn(1),0)   %the '0' here tells var to give sample variance
ans =
     0
>> var(randn(1),1)   %the '1' here tells var to give population variance
ans =
     0

Somehow, the sample variance is not dividing by $0 = n-1$ in this case. R returns a NaN for a scalar:
> var(rnorm(1,1))
[1] NA

What do you think is a sensible way to define the population sample variance for a scalar? What consequences might there be for returning a zero instead of a NaN?
edit: from the help for Matlab's var: 
VAR normalizes Y by N-1 if N>1, where N is the sample size.  This is
an unbiased estimator of the variance of the population from which X is
drawn, as long as X consists of independent, identically distributed
samples. For N=1, Y is normalized by N. 

Y = VAR(X,1) normalizes by N and produces the second moment of the
sample about its mean.  VAR(X,0) is the same as VAR(X).

a cryptic comment in the m code for `var states:
if w == 0 && n > 1
    % The unbiased estimator: divide by (n-1).  Can't do this
    % when n == 0 or 1.
    denom = n - 1;
else
    % The biased estimator: divide by n.
    denom = n; % n==0 => return NaNs, n==1 => return zeros
end

i.e. they explicitly choose not to return a NaN even when the user requests a sample variance on a scalar. My question is why they should choose to do this, not how.
edit: I see that I had erroneously asked about how one should define the population variance of a scalar (see strike through line above). This probably caused a lot of confusion.

A:

Scalars can't 'have' a population variance although they can be single samples from population that has a (population) variance.  If you want to estimate that then you need at least: more than one data point in the sample, another sample from the same distribution, or some prior information about the population variance by way of a model.
btw R has returned missing (NA) not NaN
is.nan(var(rnorm(1,1)))
[1] FALSE

A:

I am sure people in this forum will have better answers, here is what I think:
I think R's answer is logical. The random variable has a population variance, but it turns out that with 1 sample you don't have enough degrees of freedom to estimate sample variance i-e- you are trying to extract information that is NOT there.
Regarding Matlab's answer, I don't know how to justify 0, except that it is from the numerator.
Consequences can be bizarre. But I can think of anything else related to the estimation.

