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A single argument pattern definition applies to multiple-argument patterns?

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4

$begingroup$

Consider defining a pattern rule, such as

myFun[x_]:=x

As far as I understand Mathematica syntax, this rule means

whenever myFun appears with a single argument, replace the whole thing by the argument

Now, after the above definition, if we evaluate

myFun[x__]

x__

we see that even though the pattern x__ clearly symbolizes more than one argument, the single argument rule still gets executed!

Is this intended behavior? Maybe my syntax use is improper? How should I specify a single argument pattern rule which does not register with more-than-one argument patterns?

pattern-matching replacement rule argument-patterns

share|improve this question

asked 18 hours ago

KagaratschKagaratsch

4,73931348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr If I try to define myFun[x_] = x without the execution delay, I get the same behavior though...
  $endgroup$
  – Kagaratsch
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In the second case Set has the attribute HoldFirst resulting in the same behavior.
  $endgroup$
  – kglr
  18 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Consider defining a pattern rule, such as

myFun[x_]:=x

As far as I understand Mathematica syntax, this rule means

whenever myFun appears with a single argument, replace the whole thing by the argument

Now, after the above definition, if we evaluate

myFun[x__]

x__

we see that even though the pattern x__ clearly symbolizes more than one argument, the single argument rule still gets executed!

Is this intended behavior? Maybe my syntax use is improper? How should I specify a single argument pattern rule which does not register with more-than-one argument patterns?

pattern-matching replacement rule argument-patterns

share|improve this question

asked 18 hours ago

KagaratschKagaratsch

4,73931348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr If I try to define myFun[x_] = x without the execution delay, I get the same behavior though...
  $endgroup$
  – Kagaratsch
  18 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In the second case Set has the attribute HoldFirst resulting in the same behavior.
  $endgroup$
  – kglr
  18 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider defining a pattern rule, such as

myFun[x_]:=x

As far as I understand Mathematica syntax, this rule means

whenever myFun appears with a single argument, replace the whole thing by the argument

Now, after the above definition, if we evaluate

myFun[x__]

x__

we see that even though the pattern x__ clearly symbolizes more than one argument, the single argument rule still gets executed!

Is this intended behavior? Maybe my syntax use is improper? How should I specify a single argument pattern rule which does not register with more-than-one argument patterns?

pattern-matching replacement rule argument-patterns

share|improve this question

asked 18 hours ago

KagaratschKagaratsch

4,73931348

$endgroup$

myFun[x_]:=x

myFun[x__]

share|improve this question

asked 18 hours ago

KagaratschKagaratsch

4,73931348

share|improve this question

asked 18 hours ago

KagaratschKagaratsch

4,73931348

asked 18 hours ago

KagaratschKagaratsch

4,73931348

asked 18 hours ago

KagaratschKagaratsch

4,73931348

1 Answer
1

active

oldest

votes

6

$begingroup$

The pattern x__ is a Pattern object:

x__ //FullForm

Pattern[x,BlankSequence[]]

While the pattern object represents multiple arguments in a Rule or a function definition, it is a single object. Hence, your definition applies.

Why are you applying a function to a Pattern object, this is an unusual thing to do. Pattern objects usually appear inside of function definitions (Set or SetDelayed) or inside of rules (Rule or RuleDelayed).

share|improve this answer

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My trouble is that after the myFun[x_]:=x definition, if I try to use /.myFun[x__]->... type of substitutions, the substitution is applied to x__ instead of myFun[x__] which is mildly inconvenient. My workaround is to use /.myFun[x_,y__]->... instead.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Rule doesn't have any Hold attributes, so myFun evaluates. Typically, one works around this by using HoldPattern, e.g., /. HoldPattern[myFun[x__]] -> ....
  $endgroup$
  – Carl Woll
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a way to make Rule hold patterns by default? I tried Unprotect[Rule]; SetAttributes[Rule, HoldAll]; Protect[Rule]; and it seems to work that way, but I get some weird error messages along the way.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Changing Rule in this way is dangerous and not recommended. It will probably break some Mathematica functionality.
  $endgroup$
  – Carl Woll
  16 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

The pattern x__ is a Pattern object:

x__ //FullForm

Pattern[x,BlankSequence[]]

While the pattern object represents multiple arguments in a Rule or a function definition, it is a single object. Hence, your definition applies.

Why are you applying a function to a Pattern object, this is an unusual thing to do. Pattern objects usually appear inside of function definitions (Set or SetDelayed) or inside of rules (Rule or RuleDelayed).

share|improve this answer

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My trouble is that after the myFun[x_]:=x definition, if I try to use /.myFun[x__]->... type of substitutions, the substitution is applied to x__ instead of myFun[x__] which is mildly inconvenient. My workaround is to use /.myFun[x_,y__]->... instead.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Rule doesn't have any Hold attributes, so myFun evaluates. Typically, one works around this by using HoldPattern, e.g., /. HoldPattern[myFun[x__]] -> ....
  $endgroup$
  – Carl Woll
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a way to make Rule hold patterns by default? I tried Unprotect[Rule]; SetAttributes[Rule, HoldAll]; Protect[Rule]; and it seems to work that way, but I get some weird error messages along the way.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Changing Rule in this way is dangerous and not recommended. It will probably break some Mathematica functionality.
  $endgroup$
  – Carl Woll
  16 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

The pattern x__ is a Pattern object:

x__ //FullForm

Pattern[x,BlankSequence[]]

While the pattern object represents multiple arguments in a Rule or a function definition, it is a single object. Hence, your definition applies.

Why are you applying a function to a Pattern object, this is an unusual thing to do. Pattern objects usually appear inside of function definitions (Set or SetDelayed) or inside of rules (Rule or RuleDelayed).

share|improve this answer

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My trouble is that after the myFun[x_]:=x definition, if I try to use /.myFun[x__]->... type of substitutions, the substitution is applied to x__ instead of myFun[x__] which is mildly inconvenient. My workaround is to use /.myFun[x_,y__]->... instead.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Rule doesn't have any Hold attributes, so myFun evaluates. Typically, one works around this by using HoldPattern, e.g., /. HoldPattern[myFun[x__]] -> ....
  $endgroup$
  – Carl Woll
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a way to make Rule hold patterns by default? I tried Unprotect[Rule]; SetAttributes[Rule, HoldAll]; Protect[Rule]; and it seems to work that way, but I get some weird error messages along the way.
  $endgroup$
  – Kagaratsch
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kagaratsch Changing Rule in this way is dangerous and not recommended. It will probably break some Mathematica functionality.
  $endgroup$
  – Carl Woll
  16 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The pattern x__ is a Pattern object:

x__ //FullForm

Pattern[x,BlankSequence[]]

While the pattern object represents multiple arguments in a Rule or a function definition, it is a single object. Hence, your definition applies.

Why are you applying a function to a Pattern object, this is an unusual thing to do. Pattern objects usually appear inside of function definitions (Set or SetDelayed) or inside of rules (Rule or RuleDelayed).

share|improve this answer

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

$endgroup$

x__ //FullForm

share|improve this answer

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

answered 18 hours ago

Carl WollCarl Woll

70.5k394184

answered 18 hours ago

Carl WollCarl Woll

70.5k394184