Impact on website analytics caused by accessibility issues

What Does the Heart In Gyms Mean?

In iTunes 12 on macOS, how can I reset the skip count of a song?

Where is the fallacy here?

Non-Italian European mafias in USA?

Analog Mute Circuit - Simplest Solution

Is there any relevance to Thor getting his hair cut other than comedic value?

Why can't we make a perpetual motion machine by using a magnet to pull up a piece of metal, then letting it fall back down?

Is there a frame of reference in which I was born before I was conceived?

What are the issues with an additional (limited) concentration slot instead of Bladesong?

How can atoms be electrically neutral when there is a difference in the positions of the charges?

What is a term for a function that when called repeatedly, has the same effect as calling once?

Giving a talk in my old university, how prominently should I tell students my salary?

Why do members of Congress in committee hearings ask witnesses the same question multiple times?

In Adventurer's League, is it possible to keep the Ring of Winter if you manage to acquire it in the Tomb of Annihilation adventure?

Don't know what I’m looking for regarding removable HDDs?

Can throughput exceed the bandwidth of a network

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Detect if page is on experience editor Sitecore 9 via Javascript?

How can I handle a player who pre-plans arguments about my rulings on RAW?

How do ISS astronauts "get their stripes"?

What is better: yes / no radio, or simple checkbox?

Six real numbers so that product of any five is the sixth one

Is there a math equivalent to the conditional ternary operator?

Is it possible to make a clamp function shorter than a ternary in JS?

Let's create random number genratorBase-2 integer logarithm of 64-bit unsigned integerInteger square root of integerTweetable Mathematical ArtShortest Minmod FunctionNon-repeating random numbersCompute the Mertens functionMagic popcount numbersBe as evil as possibleTernary Triangles

3

$begingroup$

Imagine this short function to clamp a number between 0 and 255:

c = n => n > 0 ? n < 255 ? n : 255 : 0

Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?

P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.

code-golf math tips javascript

share|improve this question

edited 1 hour ago

Ricardo Amaral

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
  $endgroup$
  – FryAmTheEggman
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I'm well aware. I've updated the question a bit. Thank you :)
  $endgroup$
  – Ricardo Amaral
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't you at least remove all the spaces?
  $endgroup$
  – Adám
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know JS, but one way to clamp is to sort [0,n,255] and take the middle element -- might that be shorter?
  $endgroup$
  – xnor
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xnor Unfortunately, the JS sort() method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
  $endgroup$
  – Arnauld
  1 hour ago
  

  
  
  
  

 | 
show 3 more comments

3

$begingroup$

Imagine this short function to clamp a number between 0 and 255:

c = n => n > 0 ? n < 255 ? n : 255 : 0

Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?

P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.

code-golf math tips javascript

share|improve this question

edited 1 hour ago

Ricardo Amaral

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
  $endgroup$
  – FryAmTheEggman
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I'm well aware. I've updated the question a bit. Thank you :)
  $endgroup$
  – Ricardo Amaral
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't you at least remove all the spaces?
  $endgroup$
  – Adám
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know JS, but one way to clamp is to sort [0,n,255] and take the middle element -- might that be shorter?
  $endgroup$
  – xnor
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xnor Unfortunately, the JS sort() method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
  $endgroup$
  – Arnauld
  1 hour ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Imagine this short function to clamp a number between 0 and 255:

c = n => n > 0 ? n < 255 ? n : 255 : 0

Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?

P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.

code-golf math tips javascript

share|improve this question

edited 1 hour ago

Ricardo Amaral

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

c = n => n > 0 ? n < 255 ? n : 255 : 0

share|improve this question

edited 1 hour ago

Ricardo Amaral

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

Ricardo Amaral

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

New contributor

Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Ricardo AmaralRicardo Amaral

1162

1 Answer
1

active

oldest

votes

4

$begingroup$

19 bytes

You can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.

n=>n<0?0:n>>8?255:n

Try it online!

---

19 bytes

This one returns $false$ instead of $0$ but works for $nge 2^{32}$.

n=>n>255?255:n>0&&n

Try it online!

---

The original version without whitespace (and without naming the function) is 20 bytes:

n=>n>0?n<255?n:255:0

Try it online!

share|improve this answer

edited 36 mins ago

answered 1 hour ago

ArnauldArnauld

77.7k694325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go for n=>n>>8?255:n>0&&n for 18 bytes, since false can be coerced to 0 and this will make all negative numbers evaluate to false
  $endgroup$
  – Value Ink
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk I've however added another 19-byte version if supporting integers beyond 32-bit is required.
  $endgroup$
  – Arnauld
  35 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.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
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
});


}
});

Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.

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%2fcodegolf.stackexchange.com%2fquestions%2f180949%2fis-it-possible-to-make-a-clamp-function-shorter-than-a-ternary-in-js%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

19 bytes

You can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.

n=>n<0?0:n>>8?255:n

Try it online!

---

19 bytes

This one returns $false$ instead of $0$ but works for $nge 2^{32}$.

n=>n>255?255:n>0&&n

Try it online!

---

The original version without whitespace (and without naming the function) is 20 bytes:

n=>n>0?n<255?n:255:0

Try it online!

share|improve this answer

edited 36 mins ago

answered 1 hour ago

ArnauldArnauld

77.7k694325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go for n=>n>>8?255:n>0&&n for 18 bytes, since false can be coerced to 0 and this will make all negative numbers evaluate to false
  $endgroup$
  – Value Ink
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk I've however added another 19-byte version if supporting integers beyond 32-bit is required.
  $endgroup$
  – Arnauld
  35 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

19 bytes

You can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.

n=>n<0?0:n>>8?255:n

Try it online!

---

19 bytes

This one returns $false$ instead of $0$ but works for $nge 2^{32}$.

n=>n>255?255:n>0&&n

Try it online!

---

The original version without whitespace (and without naming the function) is 20 bytes:

n=>n>0?n<255?n:255:0

Try it online!

share|improve this answer

edited 36 mins ago

answered 1 hour ago

ArnauldArnauld

77.7k694325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go for n=>n>>8?255:n>0&&n for 18 bytes, since false can be coerced to 0 and this will make all negative numbers evaluate to false
  $endgroup$
  – Value Ink
  54 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk I've however added another 19-byte version if supporting integers beyond 32-bit is required.
  $endgroup$
  – Arnauld
  35 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

19 bytes

You can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.

n=>n<0?0:n>>8?255:n

Try it online!

---

19 bytes

This one returns $false$ instead of $0$ but works for $nge 2^{32}$.

n=>n>255?255:n>0&&n

Try it online!

---

The original version without whitespace (and without naming the function) is 20 bytes:

n=>n>0?n<255?n:255:0

Try it online!

share|improve this answer

edited 36 mins ago

answered 1 hour ago

ArnauldArnauld

77.7k694325

$endgroup$

n=>n<0?0:n>>8?255:n

n=>n>255?255:n>0&&n

n=>n>0?n<255?n:255:0

share|improve this answer

edited 36 mins ago

answered 1 hour ago

ArnauldArnauld

77.7k694325

answered 1 hour ago

ArnauldArnauld

77.7k694325

answered 1 hour ago

ArnauldArnauld

77.7k694325