Trjtdtk

Violin - Can double stops be played when the strings are not next to each other?

What is the adequate fee for a reveal operation?

The German vowel “a” changes to the English “i”

Bach's Toccata and Fugue in D minor breaks the "no parallel octaves" rule?

Time travel from stationary position?

Is a party consisting of only a bard, a cleric, and a warlock functional long-term?

Employee lack of ownership

Are ETF trackers fundamentally better than individual stocks?

New passport but visa is in old (lost) passport

Is it true that good novels will automatically sell themselves on Amazon (and so on) and there is no need for one to waste time promoting?

What did “the good wine” (τὸν καλὸν οἶνον) mean in John 2:10?

Welcoming 2019 Pi day: How to draw the letter π?

"of which" is correct here?

What is the Japanese sound word for the clinking of money?

How to deal with taxi scam when on vacation?

How do you talk to someone whose loved one is dying?

Are all passive ability checks floors for active ability checks?

How to pronounce "I ♥ Huckabees"?

ERC721: How to get the owned tokens of an address

Shortcut for setting origin to vertex

How do I change two letters closest to a string and one letter immediately after a string using Notepad++?

A diagram about partial derivatives of f(x,y)

Is there a place to find the pricing for things not mentioned in the PHB? (non-magical)

Do the common programs (for example: "ls", "cat") in Linux and BSD come from the same source code?

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

asked 2 hours ago

KagaratschKagaratsch

4,72931348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

add a comment | 

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

asked 2 hours ago

KagaratschKagaratsch

4,72931348

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

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 2 hours ago

KagaratschKagaratsch

4,72931348

$endgroup$

myFun[x_]:=x

myFun[x__]

share|improve this question

asked 2 hours ago

KagaratschKagaratsch

4,72931348

share|improve this question

asked 2 hours ago

KagaratschKagaratsch

4,72931348

asked 2 hours ago

KagaratschKagaratsch

4,72931348

asked 2 hours ago

KagaratschKagaratsch

4,72931348

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

answered 1 hour 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
  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
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  57 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
  42 mins ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193389%2fa-single-argument-pattern-definition-applies-to-multiple-argument-patterns%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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

answered 1 hour 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
  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
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  57 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
  42 mins ago
  

  
  
  
  

add a comment | 

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

answered 1 hour 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
  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
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, very useful, thank you!
  $endgroup$
  – Kagaratsch
  57 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
  42 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).

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

70.5k394184

$endgroup$

x__ //FullForm

share|improve this answer

answered 1 hour ago

Carl WollCarl Woll

70.5k394184

answered 1 hour ago

Carl WollCarl Woll

70.5k394184

answered 1 hour ago

Carl WollCarl Woll

70.5k394184