Is it possible to make a clamp function shorter than a ternary in JS?
$begingroup$
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
asked 3 hours ago
Ricardo AmaralRicardo Amaral
1312
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Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
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
asked 3 hours ago
Ricardo AmaralRicardo Amaral
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.
$endgroup$
$begingroup$
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 JSsort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
$endgroup$
– Arnauld
3 hours 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
asked 3 hours ago
Ricardo AmaralRicardo Amaral
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.
$endgroup$
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
asked 3 hours ago
Ricardo AmaralRicardo Amaral
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.
asked 3 hours ago
Ricardo AmaralRicardo Amaral
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.
edited 3 hours ago
Ricardo Amaral
asked 3 hours ago
Ricardo AmaralRicardo Amaral
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.
asked 3 hours ago
Ricardo AmaralRicardo Amaral
1312
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 JSsort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
$endgroup$
– Arnauld
3 hours ago
|
show 3 more comments
$begingroup$
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 JSsort()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
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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!
answered 3 hours ago
ArnauldArnauld
77.8k694325
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
3 hours ago
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will 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
add a comment |
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1 Answer
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votes
$begingroup$
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!
answered 3 hours ago
ArnauldArnauld
77.8k694325
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
3 hours ago
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will 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
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!
answered 3 hours ago
ArnauldArnauld
77.8k694325
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
3 hours ago
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will 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
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!
answered 3 hours ago
ArnauldArnauld
77.8k694325
$endgroup$
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!
answered 3 hours ago
ArnauldArnauld
77.8k694325
edited 2 hours ago
answered 3 hours ago
ArnauldArnauld
77.8k694325
answered 3 hours ago
ArnauldArnauld
77.8k694325
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 forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
3 hours ago
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will 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
add a comment |
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
3 hours ago
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will 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
add a comment |
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.
Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.
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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