Flat attribute : example I don't understand
Clash Royale CLAN TAG#URR8PPP
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]]
(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].
Hold[fonction[fonction[x]]]*)
Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?
attributes
add a comment |
I am just beginning to learn about attributes of function in mathematica.
I saw the example "Flat". But there is something I don't get :
SetAttributes[fonction, Flat]
fonction[fonction[x]]
(*fonction[x]*)
fonction[x_] := x^2;
fonction[fonction[x]]
(*$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
3
Look at the result from justfonction[x]
, you likely want to add the AttributeOneIdentity
.
– chuy
Dec 31 '18 at 17:28
add a comment |
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]]
(*$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of fonction[x].
Hold[fonction[fonction[x]]]*)
Why do I have an error ? Shouldn't it returns me fonction[x]=x^2 because of the flat attribute ?
attributes
I am just beginning to learn about attributes of function in mathematica.
I saw the example "Flat". But there is something I don't get :
SetAttributes[fonction, Flat]
fonction[fonction[x]]
(*fonction[x]*)
fonction[x_] := x^2;
fonction[fonction[x]]
(*$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
asked Dec 31 '18 at 16:38
StarBucKStarBucK
765213
765213
3
Look at the result from justfonction[x]
, you likely want to add the AttributeOneIdentity
.
– chuy
Dec 31 '18 at 17:28
add a comment |
3
Look at the result from justfonction[x]
, you likely want to add the AttributeOneIdentity
.
– 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
add a comment |
2 Answers
2
active
oldest
votes
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]
*)
add a comment |
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.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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]
*)
add a comment |
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]
*)
add a comment |
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]
*)
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]
*)
edited Jan 1 at 9:34
answered Dec 31 '18 at 17:56
Anton.SakovichAnton.Sakovich
52628
52628
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Dec 31 '18 at 17:28
Wen ChernWen Chern
35118
35118
add a comment |
add a comment |
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3
Look at the result from just
fonction[x]
, you likely want to add the AttributeOneIdentity
.– chuy
Dec 31 '18 at 17:28