Flat attribute : example I don't understand












5














I am just beginning to learn about attributes of function in mathematica.



I saw the example "Flat". But there is something I don't get :



SetAttributes[fonction, Flat]



fonction[fonction[x]]



(*fonction[x]*)



fonction[x_] := x^2;



fonction[fonction[x]]



(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].



Hold[fonction[fonction[x]]]*)


Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?






attributes







share|improve this question


















  • 3




    Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
    – chuy
    Dec 31 '18 at 17:28


















5














I am just beginning to learn about attributes of function in mathematica.



I saw the example "Flat". But there is something I don't get :



SetAttributes[fonction, Flat]



fonction[fonction[x]]



(*fonction[x]*)



fonction[x_] := x^2;



fonction[fonction[x]]



(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].



Hold[fonction[fonction[x]]]*)


Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?






attributes







share|improve this question


















  • 3




    Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
    – chuy
    Dec 31 '18 at 17:28
















5












5








5


2





I am just beginning to learn about attributes of function in mathematica.



I saw the example "Flat". But there is something I don't get :



SetAttributes[fonction, Flat]



fonction[fonction[x]]



(*fonction[x]*)



fonction[x_] := x^2;



fonction[fonction[x]]



(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].



Hold[fonction[fonction[x]]]*)


Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?






attributes







share|improve this question













I am just beginning to learn about attributes of function in mathematica.



I saw the example "Flat". But there is something I don't get :



SetAttributes[fonction, Flat]



fonction[fonction[x]]



(*fonction[x]*)



fonction[x_] := x^2;



fonction[fonction[x]]



(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].



Hold[fonction[fonction[x]]]*)


Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?






attributes




attributes






share|improve this question













share|improve this question











share|improve this question




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asked Dec 31 '18 at 16:38









StarBucKStarBucK

765213




765213








  • 3




    Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
    – chuy
    Dec 31 '18 at 17:28
















  • 3




    Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
    – chuy
    Dec 31 '18 at 17:28










3




3




Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
– chuy
Dec 31 '18 at 17:28






Look at the result from just fonction[x], you likely want to add the Attribute OneIdentity.
– chuy
Dec 31 '18 at 17:28












2 Answers
2






active

oldest

votes


















7














To understand what has happened, let's just define:



ClearAll[f];

SetAttributes[f, {Flat}];

f[x_] := Hold[x];

f[1]

(*Returns Hold[f[1]]*)


The reason for this extra f in the output is that for a Flat symbol expressions f[x] and f[f[x]] are identical. So, when a pattern-matcher encounters f[1] it treats the expression as f[f[1]] and consequently substitutes f[1], not 1, instead of x in the rhs of the definition. The pattern matcher prefers f[f[1]] over f[1] when matching x_ to allow for matching a sequence of arguments as a whole:



f[1, 2]

(*Returns Hold[f[1, 2]]*)


Here the pattern matcher treated f[1, 2] as f[f[1, 2]] and replaced x by f[1, 2] accordingly.



As chuy has already mentioned in the comments, you can add OneIdentity attribute to a symbol. Then the pattern-mathcer will prefer f[1] over f[f[1]] when matching f[x_] if there is only one argument inside the expression:



ClearAll[f];

SetAttributes[f, {Flat, OneIdentity}];

f[x_] := Hold[x];

f[1]

f[1, 2]

(*Returns Hold[1] and Hold[f[1, 2]]*)


Note, however, that OneIdentity attribute will not save your form recursion when there are more than one argument: f[1, 2] will be matched as f[f[1, 2]], f[1, 2] will be squared, f[1, 2]^2, and the f[1, 2] inside the square will again be matched as f[f[1, 2]]. So, basically, use Flat attribute only for symbols which really stand for some associative operators or you are likely to get into trouble.



ClearAll[f];

SetAttributes[f, {Flat, OneIdentity}];

f[x_] := x^2;

f[1]

f[1, 2]

(*1

  $RecursionLimit::reclim2 bla-bla-bla

  Hold[f[1, 2]^2]

*)





share|improve this answer































    3














    The following example may help:



    SetAttributes[f, Flat];
    
Hold[f[f[x]]] /. HoldPattern[f[x_]] :> x^2


    The result is:



    Hold[f[f[x]]^2]


    To see what's happening, we may run



    MatchQ[f[a, b], f[_]]


    The result is True. Thus we see that f[a,b] is identified as f[f[a,b]]. This is what the attribute Flat does to a function.






    share|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      To understand what has happened, let's just define:



      ClearAll[f];
      
SetAttributes[f, {Flat}];
      
f[x_] := Hold[x];
      
f[1]
      
(*Returns Hold[f[1]]*)


      The reason for this extra f in the output is that for a Flat symbol expressions f[x] and f[f[x]] are identical. So, when a pattern-matcher encounters f[1] it treats the expression as f[f[1]] and consequently substitutes f[1], not 1, instead of x in the rhs of the definition. The pattern matcher prefers f[f[1]] over f[1] when matching x_ to allow for matching a sequence of arguments as a whole:



      f[1, 2]
      
(*Returns Hold[f[1, 2]]*)


      Here the pattern matcher treated f[1, 2] as f[f[1, 2]] and replaced x by f[1, 2] accordingly.



      As chuy has already mentioned in the comments, you can add OneIdentity attribute to a symbol. Then the pattern-mathcer will prefer f[1] over f[f[1]] when matching f[x_] if there is only one argument inside the expression:



      ClearAll[f];
      
SetAttributes[f, {Flat, OneIdentity}];
      
f[x_] := Hold[x];
      
f[1]
      
f[1, 2]
      
(*Returns Hold[1] and Hold[f[1, 2]]*)


      Note, however, that OneIdentity attribute will not save your form recursion when there are more than one argument: f[1, 2] will be matched as f[f[1, 2]], f[1, 2] will be squared, f[1, 2]^2, and the f[1, 2] inside the square will again be matched as f[f[1, 2]]. So, basically, use Flat attribute only for symbols which really stand for some associative operators or you are likely to get into trouble.



      ClearAll[f];
      
SetAttributes[f, {Flat, OneIdentity}];
      
f[x_] := x^2;
      
f[1]
      
f[1, 2]
      
(*1
      
  $RecursionLimit::reclim2 bla-bla-bla
      
  Hold[f[1, 2]^2]
      
*)





      share|improve this answer




























        7














        To understand what has happened, let's just define:



        ClearAll[f];
        
SetAttributes[f, {Flat}];
        
f[x_] := Hold[x];
        
f[1]
        
(*Returns Hold[f[1]]*)


        The reason for this extra f in the output is that for a Flat symbol expressions f[x] and f[f[x]] are identical. So, when a pattern-matcher encounters f[1] it treats the expression as f[f[1]] and consequently substitutes f[1], not 1, instead of x in the rhs of the definition. The pattern matcher prefers f[f[1]] over f[1] when matching x_ to allow for matching a sequence of arguments as a whole:



        f[1, 2]
        
(*Returns Hold[f[1, 2]]*)


        Here the pattern matcher treated f[1, 2] as f[f[1, 2]] and replaced x by f[1, 2] accordingly.



        As chuy has already mentioned in the comments, you can add OneIdentity attribute to a symbol. Then the pattern-mathcer will prefer f[1] over f[f[1]] when matching f[x_] if there is only one argument inside the expression:



        ClearAll[f];
        
SetAttributes[f, {Flat, OneIdentity}];
        
f[x_] := Hold[x];
        
f[1]
        
f[1, 2]
        
(*Returns Hold[1] and Hold[f[1, 2]]*)


        Note, however, that OneIdentity attribute will not save your form recursion when there are more than one argument: f[1, 2] will be matched as f[f[1, 2]], f[1, 2] will be squared, f[1, 2]^2, and the f[1, 2] inside the square will again be matched as f[f[1, 2]]. So, basically, use Flat attribute only for symbols which really stand for some associative operators or you are likely to get into trouble.



        ClearAll[f];
        
SetAttributes[f, {Flat, OneIdentity}];
        
f[x_] := x^2;
        
f[1]
        
f[1, 2]
        
(*1
        
  $RecursionLimit::reclim2 bla-bla-bla
        
  Hold[f[1, 2]^2]
        
*)





        share|improve this answer


























          7












          7








          7






          To understand what has happened, let's just define:



          ClearAll[f];
          
SetAttributes[f, {Flat}];
          
f[x_] := Hold[x];
          
f[1]
          
(*Returns Hold[f[1]]*)


          The reason for this extra f in the output is that for a Flat symbol expressions f[x] and f[f[x]] are identical. So, when a pattern-matcher encounters f[1] it treats the expression as f[f[1]] and consequently substitutes f[1], not 1, instead of x in the rhs of the definition. The pattern matcher prefers f[f[1]] over f[1] when matching x_ to allow for matching a sequence of arguments as a whole:



          f[1, 2]
          
(*Returns Hold[f[1, 2]]*)


          Here the pattern matcher treated f[1, 2] as f[f[1, 2]] and replaced x by f[1, 2] accordingly.



          As chuy has already mentioned in the comments, you can add OneIdentity attribute to a symbol. Then the pattern-mathcer will prefer f[1] over f[f[1]] when matching f[x_] if there is only one argument inside the expression:



          ClearAll[f];
          
SetAttributes[f, {Flat, OneIdentity}];
          
f[x_] := Hold[x];
          
f[1]
          
f[1, 2]
          
(*Returns Hold[1] and Hold[f[1, 2]]*)


          Note, however, that OneIdentity attribute will not save your form recursion when there are more than one argument: f[1, 2] will be matched as f[f[1, 2]], f[1, 2] will be squared, f[1, 2]^2, and the f[1, 2] inside the square will again be matched as f[f[1, 2]]. So, basically, use Flat attribute only for symbols which really stand for some associative operators or you are likely to get into trouble.



          ClearAll[f];
          
SetAttributes[f, {Flat, OneIdentity}];
          
f[x_] := x^2;
          
f[1]
          
f[1, 2]
          
(*1
          
  $RecursionLimit::reclim2 bla-bla-bla
          
  Hold[f[1, 2]^2]
          
*)





          share|improve this answer














          To understand what has happened, let's just define:



          ClearAll[f];
          
SetAttributes[f, {Flat}];
          
f[x_] := Hold[x];
          
f[1]
          
(*Returns Hold[f[1]]*)


          The reason for this extra f in the output is that for a Flat symbol expressions f[x] and f[f[x]] are identical. So, when a pattern-matcher encounters f[1] it treats the expression as f[f[1]] and consequently substitutes f[1], not 1, instead of x in the rhs of the definition. The pattern matcher prefers f[f[1]] over f[1] when matching x_ to allow for matching a sequence of arguments as a whole:



          f[1, 2]
          
(*Returns Hold[f[1, 2]]*)


          Here the pattern matcher treated f[1, 2] as f[f[1, 2]] and replaced x by f[1, 2] accordingly.



          As chuy has already mentioned in the comments, you can add OneIdentity attribute to a symbol. Then the pattern-mathcer will prefer f[1] over f[f[1]] when matching f[x_] if there is only one argument inside the expression:



          ClearAll[f];
          
SetAttributes[f, {Flat, OneIdentity}];
          
f[x_] := Hold[x];
          
f[1]
          
f[1, 2]
          
(*Returns Hold[1] and Hold[f[1, 2]]*)


          Note, however, that OneIdentity attribute will not save your form recursion when there are more than one argument: f[1, 2] will be matched as f[f[1, 2]], f[1, 2] will be squared, f[1, 2]^2, and the f[1, 2] inside the square will again be matched as f[f[1, 2]]. So, basically, use Flat attribute only for symbols which really stand for some associative operators or you are likely to get into trouble.



          ClearAll[f];
          
SetAttributes[f, {Flat, OneIdentity}];
          
f[x_] := x^2;
          
f[1]
          
f[1, 2]
          
(*1
          
  $RecursionLimit::reclim2 bla-bla-bla
          
  Hold[f[1, 2]^2]
          
*)






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jan 1 at 9:34

























          answered Dec 31 '18 at 17:56









          Anton.SakovichAnton.Sakovich

          52628




          52628























              3














              The following example may help:



              SetAttributes[f, Flat];
              
Hold[f[f[x]]] /. HoldPattern[f[x_]] :> x^2


              The result is:



              Hold[f[f[x]]^2]


              To see what's happening, we may run



              MatchQ[f[a, b], f[_]]


              The result is True. Thus we see that f[a,b] is identified as f[f[a,b]]. This is what the attribute Flat does to a function.






              share|improve this answer


























                3














                The following example may help:



                SetAttributes[f, Flat];
                
Hold[f[f[x]]] /. HoldPattern[f[x_]] :> x^2


                The result is:



                Hold[f[f[x]]^2]


                To see what's happening, we may run



                MatchQ[f[a, b], f[_]]


                The result is True. Thus we see that f[a,b] is identified as f[f[a,b]]. This is what the attribute Flat does to a function.






                share|improve this answer
























                  3












                  3








                  3






                  The following example may help:



                  SetAttributes[f, Flat];
                  
Hold[f[f[x]]] /. HoldPattern[f[x_]] :> x^2


                  The result is:



                  Hold[f[f[x]]^2]


                  To see what's happening, we may run



                  MatchQ[f[a, b], f[_]]


                  The result is True. Thus we see that f[a,b] is identified as f[f[a,b]]. This is what the attribute Flat does to a function.






                  share|improve this answer












                  The following example may help:



                  SetAttributes[f, Flat];
                  
Hold[f[f[x]]] /. HoldPattern[f[x_]] :> x^2


                  The result is:



                  Hold[f[f[x]]^2]


                  To see what's happening, we may run



                  MatchQ[f[a, b], f[_]]


                  The result is True. Thus we see that f[a,b] is identified as f[f[a,b]]. This is what the attribute Flat does to a function.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 31 '18 at 17:28









                  Wen ChernWen Chern

                  35118




                  35118






























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