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I want to define a function which can take only integer arguments and have found two ways to do the same : 

func1[n_Integer, m_Integer] := If[n==m,0,n-m]
func2[n_?IntegerQ, m_?IntegerQ] := If[n==m,0,n-m]

I have also read about the difference between the two : here. Both of the above ways ensure that the functions are not evaluated if the argument is not having a Head as Integers.

However, I would like the functions to return 0 if the arguments are not integers instead of being unevaluated. One way to do this would be using a nested-if (not sure about the terminology in Mathematica) : 

func3[n_, m_] := 
 If[IntegerQ[n] && IntegerQ[m], If[n == m, 1, n - m], 0]

Is there any way other than using the above to achieve the same result? Also, if I use func1 and func2 in an integral etc such that their arguments are [5,5.2] then am I going to receive any warning of the non-evaluation or that particular value is just going to be skipped?

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    $\begingroup$ Is there any need for the If statement? Adapting the answer by bill s, maybe: f[n_?IntegerQ, m_?IntegerQ] := n-m; f[n_, m_] := 0; f[n_?IntegerQ, n_?IntegerQ] := 0; $\endgroup$
    – user1066
    May 11, 2020 at 18:36

1 Answer 1

Reset to default
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Extend the definition of your function to include all inputs:

func2[n_?IntegerQ, m_?IntegerQ] := If[n==m,0,n-m]
func2[n_, m_] := 0

Now when you input n and m integers the top definition applies, when n or m are not integers, the bottom applies. Thus func2[4, 7] is -3 while func2[54., 32] is 0. You could do the same for func1.

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  • $\begingroup$ Nice. Before I accept this as an answer, can you also point out any pitfalls of this method? Any implementation I should be careful about? Which method is better : nested-if or extending the definition ? $\endgroup$
    – Nitin
    May 11, 2020 at 18:22
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    $\begingroup$ I'm not sure it's a pitfall, but you might want to note that func2[n_,m_] as above evaluates for symbols as well as numbers. If you don't want this, then you can extend the idea to as many different definitions as make sense for your problem. $\endgroup$
    – bill s
    May 11, 2020 at 18:33
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    $\begingroup$ To avoid premature evaluation, use func2[n_?IntegerQ, m_?IntegerQ] := n - m; func2[n_, m_?(NumericQ[#] && ! IntegerQ[#] &)] = 0; func2[n_?(NumericQ[#] && ! IntegerQ[#] &), m_] = 0; $\endgroup$
    – Bob Hanlon
    May 11, 2020 at 18:57
  • $\begingroup$ Perfect. Thank you. These helped a lot. :) $\endgroup$
    – Nitin
    May 11, 2020 at 19:48

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