Testing for Null and not Null in Mathematica

Question

What is the best / cleanest / advisable way to test if a value is Null in Mathematica ? And Not Null?

For example:

 a = Null
 b = 0;
 f[n_] := If[n == Null, 1, 2]
 f[a]
 f[b]

has the result:

 1
 If[0 == Null, 1, 2]

Where I would have expected that 2 for f[b].

That does it! Equal Same. I have to check why there is a difference. — nilo de roock
I cover this topic here: mathprogramming-intro.org/book/node24.html — Leonid Shifrin

Simon Simon · Accepted Answer · 2011-06-28T22:49:29

As pointed out by Daniel (and explained in Leonid's book) Null == 0 does not evaluate to either True or False, so the If statement (as written) also does not evaluate. Null is a special Symbol that does not display in output, but in all other ways acts like a normal, everyday symbol.

In[1]:= Head[Null]
Out[1]= Symbol

For some undefined symbol x, you don't want x == 0 to return False, since x could be zero later on. This is why Null == 0 also doesn't evaluate.

There are two possible fixes for this:

1) Force the test to evaluate using TrueQ or SameQ.
For the n == Null test, the following will equivalent, but when testing numerical objects they will not. (This is because Equal uses an approximate test for numerical equivalence.)

f[n_] := If[TrueQ[n == Null], 1, 2]   (* TrueQ *)
f[n_] := If[n === Null, 1, 2]         (* SameQ *)

Using the above, the conditional statement works as you wanted:

In[3]:= {f[Null], f[0]}
Out[3]= {1, 2}

2) Use the optional 4th argument of If that is returned if the test remains unevaluated (i.e. if it is neither True nor False)

g[n_] := If[n == Null, 1, 2, 3]

Then

In[5]:= {g[Null], g[0]}
Out[5]= {1, 3}

Testing for Null and not Null in Mathematica

2 Answers