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












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
    1 hour ago






  • 1




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






  • 1




    $begingroup$
    In the second case Set has the attribute HoldFirst resulting in the same behavior.
    $endgroup$
    – kglr
    1 hour 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 2 hours ago









KagaratschKagaratsch

4,72931348




4,72931348












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






  • 1




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


















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






  • 1




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
















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




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




1




1




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




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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 hour 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
    1 hour ago










  • $begingroup$
    I see, very useful, thank you!
    $endgroup$
    – Kagaratsch
    1 hour 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
    46 mins ago











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 hour 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
    1 hour ago










  • $begingroup$
    I see, very useful, thank you!
    $endgroup$
    – Kagaratsch
    1 hour 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
    46 mins ago
















4












$begingroup$

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 hour 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
    1 hour ago










  • $begingroup$
    I see, very useful, thank you!
    $endgroup$
    – Kagaratsch
    1 hour 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
    46 mins ago














4












4








4





$begingroup$

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 1 hour ago









Carl WollCarl Woll

70.5k394184




70.5k394184












  • $begingroup$
    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 hour 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
    1 hour ago










  • $begingroup$
    I see, very useful, thank you!
    $endgroup$
    – Kagaratsch
    1 hour 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
    46 mins 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 hour 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
    1 hour ago










  • $begingroup$
    I see, very useful, thank you!
    $endgroup$
    – Kagaratsch
    1 hour 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
    46 mins 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 hour 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 hour ago




2




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
1 hour ago




$begingroup$
@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
1 hour ago












$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
1 hour ago




$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
1 hour 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
46 mins 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
46 mins ago


















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