In a notebook, I have a function that basically needs to do something like calling a pure function within a function. I would say that what best describes would be an example like:

f[x_, func_: # &] := func[x]

The problem with this is that when I use it, for example just f[x] the output is f[x] or (#1)[x] and if I call f[x, (1 + #) &] the output is (1 + #1)[x]

Even worse, if, for some reason, I decide to change the definition of f to

f[x_, func_: Exp] := func[x]

and then change it back to the original definition, the output to f[x] will now forever be Exp[x].

So, is there a way to solve this problem so that with the first definition I get the result I'm waiting f[x] = x and f[x, (1 + #) &] = 1 + x?

Thanks for the help.

  • 7
    $\begingroup$ f[x_, func_: (# &)] := func[x] otherwise it is (func_: #) & $\endgroup$
    – Kuba
    Oct 25, 2018 at 9:28
  • $\begingroup$ Yeah, that seems to solve it! Thanks $\endgroup$
    – Argidore
    Oct 25, 2018 at 9:47
  • $\begingroup$ From all the stuff that I tried to find a solution, never thought it would be as simple as that $\endgroup$
    – Argidore
    Oct 25, 2018 at 9:50
  • $\begingroup$ @Argidore There exists a concept Precedence of operators. The problem arises from the fact that Function (&) has a rather low precedence. This can be seen by running this command: Precedence /@ {Pattern, Optional, Function}. $\endgroup$ Oct 25, 2018 at 9:56
  • $\begingroup$ Arn't the two definitions f[x_, func_: (# &)] := func[x] and F[x_, func_] := func[x] the same? $\endgroup$ Oct 25, 2018 at 14:19

1 Answer 1


To specify values to a function's argument(s) which take effect when calling the function but no explicit values are going to be set to them, besides Optional (:), one can also use Default (.):

Default[f, 2] = #^2 &;

f[x_, func_.] := func[x]


In this case, no bother with precedences of operators.


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