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


How to Enable Syntax Coloring of Pattern Match Variable Only (i.e. Without Coloring any Associated Pattern)?Deep Pattern matching with repeating argumentsHow to apply multiple/complicated requirements for a pattern in a function inputDetecting a more general patternHow do I tell pattern searcher the order in which to search for patterns given in general form?Confused by the opts : OptionsPattern[ ] patternHow does one unpack the contents of a list while using rules to substitute values for each variable?Combinations of multiple matching patternsLogical AND of multiple patternsOptimize DownValues: extract “non-patterns” from Alternatives













2












$begingroup$


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









$endgroup$











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






  • 1




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















2












$begingroup$


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









$endgroup$











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






  • 1




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













2












2








2





$begingroup$


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









$endgroup$




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




pattern-matching replacement rule argument-patterns






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 1 hour ago









KagaratschKagaratsch

4,72731348




4,72731348











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






  • 1




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
















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






  • 1




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















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




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




1




1




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




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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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









$endgroup$












  • $begingroup$
    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
    1 min ago










Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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









$endgroup$












  • $begingroup$
    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
    1 min ago















3












$begingroup$

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









$endgroup$












  • $begingroup$
    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
    1 min ago













3












3








3





$begingroup$

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









$endgroup$



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












share|improve this answer



share|improve this answer










answered 49 mins ago









Carl WollCarl Woll

70.4k394184




70.4k394184











  • $begingroup$
    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
    1 min ago
















  • $begingroup$
    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
    1 min ago















$begingroup$
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
1 min ago




$begingroup$
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
1 min ago

















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