As x goes to 0, the limits of x^0 and 0^x go to 1 and 0, respectively. Some say that the best value for 0^0 depends on context, others say that the value of 0^0 should be 1. I'm interested in knowing what your language implementers say. There doesn't seem to be a consensus. For example, in ActiveState Perl 5.12.0:

C:\>perl -e "print 0**0"

While in Mathematica 6.0:

In[1]:= 0^0
        During evaluation of In[1]:= Power::indet:
        Indeterminate expression 0^0 encountered. >>
Out[1]= Indeterminate

I'm also interested in knowing your opinion on the matter: What should 0^0 be? And also whether you have ever introduced a bug into a program you were writing (or had a hard time debugging a program) because of the way your language handles indeterminate forms?

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    What is the actual question? There is consensus as to what 0^0 means; it's indeterminate. Languages that don't flag it as such are wrong. What's to discuss? It's no different than if a language returned anything other than undefined for 1/0. – user8 Oct 5 '10 at 23:27
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    @Mark: actually, for 1/0 it can return 'infinity', if it has such a thing available. A better example would be '0/0', which really is indeterminate. – Jerry Coffin Oct 6 '10 at 0:11
  • Jerry: fair enough. I was thinking more of the "division by zero" assertion/exception/error most languages have that would produce an undefined value. Such behavior is more or less uncontroversial; 0^0 should be, too. – user8 Oct 6 '10 at 0:15
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    @Mark: I don't agree that there's a consensus. According to Donald Knuth, 0^0 "has to be 1." August Mobius agrees that its 1. See en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power – A. N. Other Oct 6 '10 at 1:28
  • @Mark: There are three related questions. Let me restate them succinctly: 1- Which languages implement 0^0=1 and which implement it as "Indeterminate" (or something else?) 2- Does it matter to you which way its implemented? 3- Has the way its implemented ever caused you practical issues? – A. N. Other Oct 6 '10 at 1:32

According to this Wikipedia article,

"Most programming language with a power function are implemented using the IEEE pow function and therefore evaluate 0^0 as 1. The later C and C++ standards describe this as the normative behavior. The Java standard mandates this behavior. The .NET Framework method System.Math.Pow also treats 0^0 as 1."

  • Good insight. So clearly, Mathematica isn't IEEE-compliant, since it returns 0^0=Indeterminate rather than 0^0=1. Perhaps, then, the first part of my question should be rephrased as, "Which programming languages aren't IEEE-compliant, and return something other than 1 for 0^0?" – A. N. Other Oct 6 '10 at 3:04


>>> for t in [int, float, complex, fractions.Fraction, decimal.Decimal]:
...     print(t, t(0)**0)
<type 'int'> 1
<type 'float'> 1.0
<type 'complex'> (1+0j)
<class 'fractions.Fraction'> 1
Traceback (most recent call last):
decimal.InvalidOperation: 0 ** 0
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    This is very, very interesting: So, in Python, the value of 0^0 depends on the numeric type of the base! Bravo and +1 for taking the trouble to find this out!!! – A. N. Other Oct 6 '10 at 21:01
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    Elisha: well, it also depends on each type's intended purpose. Most of the above types follow the convention to return 1, as seems generally more convenient. The decimal module/type, on the other hand, is intended to be a standards-compliant implementation of the IBM/IEEE decimal floating point specifications, which might be more conservative about 0**0. (I'm not sure, though.) – Pi Delport Oct 6 '10 at 22:01

Google says:

0^0 = 1


Delphi's got 2 float values: NaN and Infinity. That get returned in case of weirdness.

But it doesn't have a built in exponentiation function.

^ is for pointers, as in "hey you up there, I'm pointing at you!"

  • I always heard ^ pronounced "hat". – Jesse C. Slicer Oct 6 '10 at 14:14
  • Makes more sense than carrot, or carat... – Peter Turner Oct 6 '10 at 14:37
  • Standard Pascal has no exponentiation operator either. Timo Salmi contributed an exponentiation function to the Pascal FAQ here midnightbeach.com/jon/pubs/clp-faq.htm#exponentiation which looks as if it makes 0^0=1 rather than NaN. Do you agree with this implementation? – A. N. Other Oct 6 '10 at 21:21

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