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












6












$begingroup$


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.
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  • $begingroup$
    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
    3 hours ago










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










  • $begingroup$
    Shouldn't you at least remove all the spaces?
    $endgroup$
    – Adám
    3 hours 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
    3 hours 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
    3 hours ago
















6












$begingroup$


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.







$endgroup$












  • $begingroup$
    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
    3 hours ago










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










  • $begingroup$
    Shouldn't you at least remove all the spaces?
    $endgroup$
    – Adám
    3 hours 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
    3 hours 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
    3 hours ago














6












6








6


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 3 hours ago







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 3 hours ago









Ricardo AmaralRicardo Amaral

1312




1312




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.












  • $begingroup$
    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
    3 hours ago










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










  • $begingroup$
    Shouldn't you at least remove all the spaces?
    $endgroup$
    – Adám
    3 hours 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
    3 hours 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
    3 hours ago


















  • $begingroup$
    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
    3 hours ago










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










  • $begingroup$
    Shouldn't you at least remove all the spaces?
    $endgroup$
    – Adám
    3 hours 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
    3 hours 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
    3 hours ago
















$begingroup$
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
3 hours ago




$begingroup$
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
3 hours ago












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




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












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




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




1




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
3 hours ago




$begingroup$
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
3 hours ago




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
3 hours ago




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










1 Answer
1






active

oldest

votes


















7












$begingroup$

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











$endgroup$













  • $begingroup$
    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
    3 hours ago










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










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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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











$endgroup$













  • $begingroup$
    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
    3 hours ago










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










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
















7












$begingroup$

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











$endgroup$













  • $begingroup$
    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
    3 hours ago










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










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














7












7








7





$begingroup$

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











$endgroup$



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














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 3 hours ago









ArnauldArnauld

77.8k694325




77.8k694325












  • $begingroup$
    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
    3 hours ago










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










  • $begingroup$
    @ValueInk I've however added another 19-byte version if supporting integers beyond 32-bit is required.
    $endgroup$
    – Arnauld
    2 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
    3 hours ago










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










  • $begingroup$
    @ValueInk I've however added another 19-byte version if supporting integers beyond 32-bit is required.
    $endgroup$
    – Arnauld
    2 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
3 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
3 hours ago












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




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












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




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










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










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