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

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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
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
@kglr If I try to definemyFun[x_] = x
without the execution delay, I get the same behavior though...
$endgroup$
– Kagaratsch
1 hour ago
1
$begingroup$
In the second caseSet
has the attributeHoldFirst
resulting in the same behavior.
$endgroup$
– kglr
1 hour 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
$endgroup$
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
asked 1 hour ago
KagaratschKagaratsch
4,72931348
4,72931348
$begingroup$
@kglr If I try to definemyFun[x_] = x
without the execution delay, I get the same behavior though...
$endgroup$
– Kagaratsch
1 hour ago
1
$begingroup$
In the second caseSet
has the attributeHoldFirst
resulting in the same behavior.
$endgroup$
– kglr
1 hour ago
add a comment |
$begingroup$
@kglr If I try to definemyFun[x_] = x
without the execution delay, I get the same behavior though...
$endgroup$
– Kagaratsch
1 hour ago
1
$begingroup$
In the second caseSet
has the attributeHoldFirst
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
).
$endgroup$
$begingroup$
My trouble is that after themyFun[x_]:=x
definition, if I try to use/.myFun[x__]->...
type of substitutions, the substitution is applied tox__
instead ofmyFun[x__]
which is mildly inconvenient. My workaround is to use/.myFun[x_,y__]->...
instead.
$endgroup$
– Kagaratsch
12 mins ago
1
$begingroup$
@KagaratschRule
doesn't have any Hold attributes, somyFun
evaluates. Typically, one works around this by using HoldPattern, e.g.,/. HoldPattern[myFun[x__]] -> ...
.
$endgroup$
– Carl Woll
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to makeRule
hold patterns by default?
$endgroup$
– Kagaratsch
7 mins ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
).
$endgroup$
$begingroup$
My trouble is that after themyFun[x_]:=x
definition, if I try to use/.myFun[x__]->...
type of substitutions, the substitution is applied tox__
instead ofmyFun[x__]
which is mildly inconvenient. My workaround is to use/.myFun[x_,y__]->...
instead.
$endgroup$
– Kagaratsch
12 mins ago
1
$begingroup$
@KagaratschRule
doesn't have any Hold attributes, somyFun
evaluates. Typically, one works around this by using HoldPattern, e.g.,/. HoldPattern[myFun[x__]] -> ...
.
$endgroup$
– Carl Woll
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to makeRule
hold patterns by default?
$endgroup$
– Kagaratsch
7 mins 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
).
$endgroup$
$begingroup$
My trouble is that after themyFun[x_]:=x
definition, if I try to use/.myFun[x__]->...
type of substitutions, the substitution is applied tox__
instead ofmyFun[x__]
which is mildly inconvenient. My workaround is to use/.myFun[x_,y__]->...
instead.
$endgroup$
– Kagaratsch
12 mins ago
1
$begingroup$
@KagaratschRule
doesn't have any Hold attributes, somyFun
evaluates. Typically, one works around this by using HoldPattern, e.g.,/. HoldPattern[myFun[x__]] -> ...
.
$endgroup$
– Carl Woll
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to makeRule
hold patterns by default?
$endgroup$
– Kagaratsch
7 mins 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
).
$endgroup$
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
).
answered 1 hour ago


Carl WollCarl Woll
70.5k394184
70.5k394184
$begingroup$
My trouble is that after themyFun[x_]:=x
definition, if I try to use/.myFun[x__]->...
type of substitutions, the substitution is applied tox__
instead ofmyFun[x__]
which is mildly inconvenient. My workaround is to use/.myFun[x_,y__]->...
instead.
$endgroup$
– Kagaratsch
12 mins ago
1
$begingroup$
@KagaratschRule
doesn't have any Hold attributes, somyFun
evaluates. Typically, one works around this by using HoldPattern, e.g.,/. HoldPattern[myFun[x__]] -> ...
.
$endgroup$
– Carl Woll
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to makeRule
hold patterns by default?
$endgroup$
– Kagaratsch
7 mins ago
add a comment |
$begingroup$
My trouble is that after themyFun[x_]:=x
definition, if I try to use/.myFun[x__]->...
type of substitutions, the substitution is applied tox__
instead ofmyFun[x__]
which is mildly inconvenient. My workaround is to use/.myFun[x_,y__]->...
instead.
$endgroup$
– Kagaratsch
12 mins ago
1
$begingroup$
@KagaratschRule
doesn't have any Hold attributes, somyFun
evaluates. Typically, one works around this by using HoldPattern, e.g.,/. HoldPattern[myFun[x__]] -> ...
.
$endgroup$
– Carl Woll
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to makeRule
hold patterns by default?
$endgroup$
– Kagaratsch
7 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
12 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
12 mins ago
1
1
$begingroup$
@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
11 mins 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
11 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
I see, very useful, thank you!
$endgroup$
– Kagaratsch
9 mins ago
$begingroup$
Is there a way to make
Rule
hold patterns by default?$endgroup$
– Kagaratsch
7 mins ago
$begingroup$
Is there a way to make
Rule
hold patterns by default?$endgroup$
– Kagaratsch
7 mins ago
add a comment |
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qb51H4djFE23 lT
$begingroup$
@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 attributeHoldFirst
resulting in the same behavior.$endgroup$
– kglr
1 hour ago