Is it possible to make a clamp function shorter than a ternary in JS?Let's create random number...

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


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19












$begingroup$


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









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.







$endgroup$








  • 2




    $begingroup$
    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
    yesterday






  • 1




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






  • 2




    $begingroup$
    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
    yesterday






  • 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
    yesterday






  • 5




    $begingroup$
    @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
    $endgroup$
    – xnor
    yesterday
















19












$begingroup$


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









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.







$endgroup$








  • 2




    $begingroup$
    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
    yesterday






  • 1




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






  • 2




    $begingroup$
    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
    yesterday






  • 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
    yesterday






  • 5




    $begingroup$
    @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
    $endgroup$
    – xnor
    yesterday














19












19








19


1



$begingroup$


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









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.







$endgroup$




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




code-golf math tips javascript






share|improve this question









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









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




share|improve this question








edited yesterday







Ricardo Amaral













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 yesterday









Ricardo AmaralRicardo Amaral

1986




1986




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.





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.






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








  • 2




    $begingroup$
    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
    yesterday






  • 1




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






  • 2




    $begingroup$
    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
    yesterday






  • 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
    yesterday






  • 5




    $begingroup$
    @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
    $endgroup$
    – xnor
    yesterday














  • 2




    $begingroup$
    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
    yesterday






  • 1




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






  • 2




    $begingroup$
    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
    yesterday






  • 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
    yesterday






  • 5




    $begingroup$
    @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
    $endgroup$
    – xnor
    yesterday








2




2




$begingroup$
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
yesterday




$begingroup$
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
yesterday




1




1




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




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




2




2




$begingroup$
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
yesterday




$begingroup$
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
yesterday




1




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
yesterday




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




5




5




$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday




$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday










1 Answer
1






active

oldest

votes


















20












$begingroup$

20 bytes



For reference, this is the original version without whitespace and without naming the function:





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


Try it online!




19 bytes



We 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!




18 bytes



By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.





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


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    n > 0 &&  //   return false if n is negative

    255       //   or 255 otherwise

  :           // else:

    n         //   return n unchanged


(This is a fixed revision of the code proposed by @ValueInk in the comments.)




17 bytes



We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:





n=>n>>8?-n>>>24:n


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    -n >>> 24 //   non-arithmetic right-shift of -n by 24 positions

  :           // else:

    n         //   return n unchanged





share|improve this answer











$endgroup$













  • $begingroup$
    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
    yesterday










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










  • $begingroup$
    Very nice, thank you so much!
    $endgroup$
    – Ricardo Amaral
    20 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









20












$begingroup$

20 bytes



For reference, this is the original version without whitespace and without naming the function:





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


Try it online!




19 bytes



We 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!




18 bytes



By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.





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


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    n > 0 &&  //   return false if n is negative

    255       //   or 255 otherwise

  :           // else:

    n         //   return n unchanged


(This is a fixed revision of the code proposed by @ValueInk in the comments.)




17 bytes



We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:





n=>n>>8?-n>>>24:n


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    -n >>> 24 //   non-arithmetic right-shift of -n by 24 positions

  :           // else:

    n         //   return n unchanged





share|improve this answer











$endgroup$













  • $begingroup$
    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
    yesterday










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










  • $begingroup$
    Very nice, thank you so much!
    $endgroup$
    – Ricardo Amaral
    20 hours ago
















20












$begingroup$

20 bytes



For reference, this is the original version without whitespace and without naming the function:





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


Try it online!




19 bytes



We 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!




18 bytes



By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.





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


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    n > 0 &&  //   return false if n is negative

    255       //   or 255 otherwise

  :           // else:

    n         //   return n unchanged


(This is a fixed revision of the code proposed by @ValueInk in the comments.)




17 bytes



We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:





n=>n>>8?-n>>>24:n


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    -n >>> 24 //   non-arithmetic right-shift of -n by 24 positions

  :           // else:

    n         //   return n unchanged





share|improve this answer











$endgroup$













  • $begingroup$
    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
    yesterday










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










  • $begingroup$
    Very nice, thank you so much!
    $endgroup$
    – Ricardo Amaral
    20 hours ago














20












20








20





$begingroup$

20 bytes



For reference, this is the original version without whitespace and without naming the function:





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


Try it online!




19 bytes



We 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!




18 bytes



By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.





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


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    n > 0 &&  //   return false if n is negative

    255       //   or 255 otherwise

  :           // else:

    n         //   return n unchanged


(This is a fixed revision of the code proposed by @ValueInk in the comments.)




17 bytes



We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:





n=>n>>8?-n>>>24:n


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    -n >>> 24 //   non-arithmetic right-shift of -n by 24 positions

  :           // else:

    n         //   return n unchanged





share|improve this answer











$endgroup$



20 bytes



For reference, this is the original version without whitespace and without naming the function:





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


Try it online!




19 bytes



We 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!




18 bytes



By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.





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


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    n > 0 &&  //   return false if n is negative

    255       //   or 255 otherwise

  :           // else:

    n         //   return n unchanged


(This is a fixed revision of the code proposed by @ValueInk in the comments.)




17 bytes



We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:





n=>n>>8?-n>>>24:n


Try it online!



Commented



n =>          // n = input number

  n >> 8 ?    // if n is greater than 255 or n is negative:

    -n >>> 24 //   non-arithmetic right-shift of -n by 24 positions

  :           // else:

    n         //   return n unchanged






share|improve this answer














share|improve this answer



share|improve this answer








edited 21 hours ago

























answered yesterday









ArnauldArnauld

77.9k695326




77.9k695326












  • $begingroup$
    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
    yesterday










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










  • $begingroup$
    Very nice, thank you so much!
    $endgroup$
    – Ricardo Amaral
    20 hours ago


















  • $begingroup$
    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
    yesterday










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










  • $begingroup$
    Very nice, thank you so much!
    $endgroup$
    – Ricardo Amaral
    20 hours ago
















$begingroup$
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
yesterday




$begingroup$
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
yesterday












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




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












$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago




$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago










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










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













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












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
















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