Inactivate all occurrences of a symbol

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Hi crowd of knowledge,



I would like to inactivate all occurrences of a symbol to prevent it from evaluating, regardless where in the evaluation process it appears. A minimal example:



g[x_] := foo;
f[x_] := x + g[x];
Inactivate[f[2] + g[3], g]
(*== output ==*)
2 + f00 + Inactive[g][3]


My desired output would be 2 + Inactive[g][2] + Inactive[g][3].



I understand that there would be thorny issues in general how to handle scoping constructs such as Block and Module but my naïve wish is "inactivate everywhere there is a head g and don't care about anything else".



(Additional comments: I don't want to write a temporary overwriting definition of g as the function of my actual interest has several down values which take quite some time to compute. Something like [the fictitious] SetAttributes[g,NeverEvaluates] could be something though.)






evaluation







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    up vote
    2
    down vote

    favorite












    Hi crowd of knowledge,



    I would like to inactivate all occurrences of a symbol to prevent it from evaluating, regardless where in the evaluation process it appears. A minimal example:



    g[x_] := foo;
    f[x_] := x + g[x];
    Inactivate[f[2] + g[3], g]
    (*== output ==*)
    2 + f00 + Inactive[g][3]


    My desired output would be 2 + Inactive[g][2] + Inactive[g][3].



    I understand that there would be thorny issues in general how to handle scoping constructs such as Block and Module but my naïve wish is "inactivate everywhere there is a head g and don't care about anything else".



    (Additional comments: I don't want to write a temporary overwriting definition of g as the function of my actual interest has several down values which take quite some time to compute. Something like [the fictitious] SetAttributes[g,NeverEvaluates] could be something though.)






    evaluation







    share|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Hi crowd of knowledge,



      I would like to inactivate all occurrences of a symbol to prevent it from evaluating, regardless where in the evaluation process it appears. A minimal example:



      g[x_] := foo;
      f[x_] := x + g[x];
      Inactivate[f[2] + g[3], g]
      (*== output ==*)
      2 + f00 + Inactive[g][3]


      My desired output would be 2 + Inactive[g][2] + Inactive[g][3].



      I understand that there would be thorny issues in general how to handle scoping constructs such as Block and Module but my naïve wish is "inactivate everywhere there is a head g and don't care about anything else".



      (Additional comments: I don't want to write a temporary overwriting definition of g as the function of my actual interest has several down values which take quite some time to compute. Something like [the fictitious] SetAttributes[g,NeverEvaluates] could be something though.)






      evaluation







      share|improve this question













      Hi crowd of knowledge,



      I would like to inactivate all occurrences of a symbol to prevent it from evaluating, regardless where in the evaluation process it appears. A minimal example:



      g[x_] := foo;
      f[x_] := x + g[x];
      Inactivate[f[2] + g[3], g]
      (*== output ==*)
      2 + f00 + Inactive[g][3]


      My desired output would be 2 + Inactive[g][2] + Inactive[g][3].



      I understand that there would be thorny issues in general how to handle scoping constructs such as Block and Module but my naïve wish is "inactivate everywhere there is a head g and don't care about anything else".



      (Additional comments: I don't want to write a temporary overwriting definition of g as the function of my actual interest has several down values which take quite some time to compute. Something like [the fictitious] SetAttributes[g,NeverEvaluates] could be something though.)






      evaluation




      evaluation






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      Stalpotaten

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          I believe that this works, your final paragraph notwithstanding as the downvalues of g are not lost by using Block.



          Block[g = Inactive[g], f[2] + g[3]]



          2 + Inactive[g][2] + Inactive[g][3]



          Since Inactive has HoldFirst infinite recursion does not occur.






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            up vote
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            I believe that this works, your final paragraph notwithstanding as the downvalues of g are not lost by using Block.



            Block[g = Inactive[g], f[2] + g[3]]



            2 + Inactive[g][2] + Inactive[g][3]



            Since Inactive has HoldFirst infinite recursion does not occur.






            share|improve this answer


























              up vote
              4
              down vote













              I believe that this works, your final paragraph notwithstanding as the downvalues of g are not lost by using Block.



              Block[g = Inactive[g], f[2] + g[3]]



              2 + Inactive[g][2] + Inactive[g][3]



              Since Inactive has HoldFirst infinite recursion does not occur.






              share|improve this answer
























                up vote
                4
                down vote










                up vote
                4
                down vote









                I believe that this works, your final paragraph notwithstanding as the downvalues of g are not lost by using Block.



                Block[g = Inactive[g], f[2] + g[3]]



                2 + Inactive[g][2] + Inactive[g][3]



                Since Inactive has HoldFirst infinite recursion does not occur.






                share|improve this answer














                I believe that this works, your final paragraph notwithstanding as the downvalues of g are not lost by using Block.



                Block[g = Inactive[g], f[2] + g[3]]



                2 + Inactive[g][2] + Inactive[g][3]



                Since Inactive has HoldFirst infinite recursion does not occur.







                share|improve this answer














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                share|improve this answer








                edited 12 mins ago

























                answered 1 hour ago









                Mr.Wizard♦

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